NYOJ 搜索题目汇总 NYOJ 20、21、27、42、58、82、202、284、325、353、488、491、523、592、722

NYOJ 搜索题目汇总

NYOJ 20

 #include<iostream>#include<cstdio>#include<cstring>#include<string>#include<vector>#include<algorithm>using namespace std;int pre[100005];vector<int>v[100005];//存储每个结点相邻的边void DFS(int cur){for(int i = 0; i < v[cur].size(); ++i)//边{if(pre[v[cur][i]]) continue; //若存在父节点则继续遍历pre[v[cur][i]] = cur; //相连节点的父节点为curDFS(v[cur][i]); //深搜到底，把一条路上父节点全部找出}}int main(){int ncase, num, cur, i, x, y;scanf("%d", &ncase);while(ncase--){memset(v, 0, sizeof(v));memset(pre, 0, sizeof(pre));scanf("%d %d", &num, &cur);pre[cur] = - 1; //起点没有父节点for(i = 0; i < num - 1; ++i){scanf("%d %d", &x, &y);v[x].push_back(y); //x与y相连v[y].push_back(x); //y与x也肯定相连}DFS(cur); //起点开始深搜for(i = 1; i <= num; ++i)printf("%d ", pre[i]); //每个节点的父节点都保存在pri数组，输出即可}return 0;}        

NYOJ 21

法一：

 #include<iostream>#include<cstring>#include<cstdio>#include<queue>using namespace std;int total;int v1, v2, v3;bool visit[105][105][105]; //状态是否出现struct state{int a, b, c;int ceng; //最小步数state(){a = b = c = ceng = 0;}}b, e;int BFS(state begin){queue<state> q;while(!q.empty())q.pop();q.push(b);while(!q.empty()){state cur = q.front();q.pop();visit[cur.a][cur.b][cur.c] = true;if(cur.a == e.a && cur.b == e.b && cur.c == e.c) //找到return cur.ceng;if(cur.a > 0 && cur.b < v2)//v1->v2{state temp = cur;int tt = min(temp.a, v2 - temp.b);temp.a -= tt;temp.b += tt;if(visit[temp.a][temp.b][temp.c] == false){visit[temp.a][temp.b][temp.c] = true;temp.ceng++;q.push(temp);}}if(cur.a > 0 && cur.c < v3)//v1->v3{state temp = cur;int tt = min(temp.a, v3 - temp.c);temp.a -= tt;temp.c += tt;if(visit[temp.a][temp.b][temp.c] == false){visit[temp.a][temp.b][temp.c] = true;temp.ceng++;q.push(temp);}}if(cur.b > 0 && cur.a < v1)//v2->v1{state temp = cur;int tt = min(temp.b, v1 - temp.a);temp.a += tt;temp.b -= tt;if(visit[temp.a][temp.b][temp.c] == false){visit[temp.a][temp.b][temp.c] = true;temp.ceng++;q.push(temp);}}if(cur.b > 0 && cur.c < v3)//v2->v3{state temp = cur;int tt = min(temp.b, v3 - temp.c);temp.b -= tt;temp.c += tt;if(visit[temp.a][temp.b][temp.c] == false){visit[temp.a][temp.b][temp.c] = true;temp.ceng++;q.push(temp);}}if(cur.c > 0 && cur.a < v1)//v3->v1{state temp = cur;int tt = min(temp.c, v1 - temp.a);temp.c -= tt;temp.a += tt;if(visit[temp.a][temp.b][temp.c] == false){visit[temp.a][temp.b][temp.c] = true;temp.ceng++;q.push(temp);}}if(cur.c > 0 && cur.b < v2)//v3->v2{state temp = cur;int tt = min(temp.c, v2 - temp.b);temp.c -= tt;temp.b += tt;if(visit[temp.a][temp.b][temp.c] == false){visit[temp.a][temp.b][temp.c] = true;temp.ceng++;q.push(temp);}}}return -1; //没有终状态}int main(){int ncase;scanf("%d", &ncase);while(ncase--){total = 0;memset(visit, false, sizeof(visit));scanf("%d %d %d", &v1, &v2, &v3);b.a = v1, b.b = 0, b.c = 0, b.ceng = 0;scanf("%d %d %d", &e.a, &e.b, &e.c);if(v1 < e.a + e.b + e.c){printf("-1\n");continue;}elseprintf("%d\n", BFS(b));}return 0;}        

法二：

 #include <stdio.h>#include <string.h>#define MOVE(x) (x=(x+1)%1000)bool state[100][100];struct Size{int e[3];}queue[1000];int V[3],E1,E2,E3;int BFS(){if(E1+E2+E3!=V[0])return -1;else if(E1<0 || E2<0 || E3<0)return -1;else if(V[0]==E1 && 0==E2 && 0==E3)return 0;elsememset(state,false,sizeof(state));int tpe,cnt=0;int front,rear;int num,temp;Size s;front=rear=-1; s.e[0]=V[0]; s.e[1]=0; s.e[2]=0;queue[MOVE(rear)]=s; state[s.e[0]][s.e[1]]=true; num=1; temp=0;   while(front!=rear){s=queue[MOVE(front)];if(--num==0)cnt++;for(int i=0;i<3;i++){if(s.e[i]){for(int j=(i+1)%3;j!=i;j=(j+1)%3){tpe=(s.e[i]>V[j]-s.e[j])?(V[j]-s.e[j]):s.e[i];s.e[i]-=tpe;s.e[j]+=tpe;if(s.e[0]==E1 && s.e[1]==E2 && s.e[2]==E3)return num?cnt+1:cnt;if(!state[s.e[0]][s.e[1]]){queue[MOVE(rear)]=s;state[s.e[0]][s.e[1]]=true;temp++;}s.e[i]+=tpe;s.e[j]-=tpe;}}}if(num==0)num=temp,temp=0;}return -1;}int main(){int t;scanf("%d",&t);while(t--){scanf("%d %d %d",&V[0],&V[1],&V[2]);scanf("%d %d %d",&E1,&E2,&E3);printf("%d\n",BFS());}return 0;}        

NYOJ 27

 #include<iostream>#include<cstring>const int N=100;using namespace std;int n,m;int mp[N][N];int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};void dfs(int si,int sj){mp[si][sj] = 0;int di,i,j;for(di=0;di<4;di++){i = si + dir[di][0];j = sj + dir[di][1];if (i>=0 &&i<m && j>=0 && j<n && mp[i][j])dfs(i,j);}}int main(){int t,cnt;cin>>t;while(t--){cin>>m>>n;int i,j;for(i=0;i<m;i++)for(j=0;j<n;j++)cin>>mp[i][j];cnt = 0;for(i=0;i<m;i++)for(j=0;j<n;j++)if (mp[i][j]==1){cnt++;dfs(i,j);}cout<<cnt<<endl;}return 0;}        

NYOJ 42

#include<stdio.h>#include<string.h>int map[1050][1050],vis[1050],chu[1050];int P,Q;void dfs(int point){    int i;    vis[point]=1;//DFS 深度遍历    for(i=1;i<=P;i++)        if(map[point][i]!=0)        {           chu[point]++;           if(vis[i]==0)            dfs(i);        }}int main(){    int i,j,N;    scanf("%d",&N);    while(N--)    {        int a,b;        int all_vis=1;        memset(map,0,sizeof(map));        memset(vis,0,sizeof(vis));        memset(chu,0,sizeof(chu));        scanf("%d %d",&P,&Q);        for(i=0;i<Q;i++)        {            scanf("%d %d",&a,&b);            map[a][b]=1;            map[b][a]=1;        }        dfs(1);        for(i=1;i<P;i++)        {            if(vis[i]==0)            {                all_vis=0;                break;            }        }        if(all_vis==0) printf("No\n");        else        {            j=0;            for(i=1;i<=P;i++)                if(chu[i]%2==1) j+=1;//欧拉图 度为奇数            if(j==0||j==2) printf("Yes\n");            else printf("No\n");        }    }}

NYOJ 58

#include<stdio.h>int a[9][9]={1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,0,1,1,0,0,1,1,0,0,0,1,1,0,1,0,1,1,0,1,1,1,0,0,0,0,1,0,0,1,1,1,0,1,0,1,0,0,1,1,1,0,1,0,1,0,0,1,1,1,0,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1};int t,x1,x2,y1,y2,m;int dfs(int x,int y,int s){    if(x==x2&&y==y2){if(m>s)m=s;}    else        {        if(!a[x+1][y])  {a[x+1][y]=1;dfs(x+1,y,s+1);a[x+1][y]=0;}        if(!a[x-1][y])  {a[x-1][y]=1;dfs(x-1,y,s+1);a[x-1][y]=0;}        if(!a[x][y+1])  {a[x][y+1]=1;dfs(x,y+1,s+1);a[x][y+1]=0;}        if(!a[x][y-1])  {a[x][y-1]=1;dfs(x,y-1,s+1);a[x][y-1]=0;}    }}int main(){    int s;    scanf("%d",&t);    while(t--)    {        m=81;s=0;        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);        dfs(x1,y1,s);        printf("%d\n",m);    }return 0;}

NYOJ 82

法一：

#include <iostream>#include <queue>#include <cstdio>#include <cstdlib>using namespace std;typedef struct pos{   int x,y;} POS;int d[4][2]={{-1,0},{0,1},{1,0},{0,-1}};int bfs(POS from);char maze[20][20];int key[5];queue<pos> que;int row,col;POS start;int main(){   int i,j;     while(scanf("%d%d",&row,&col) && (row + col))     {         for(i = 0 ; i < 5 ; ++i)         key[i] = 0;         for(i = 0 ; i < row ; ++i)         {               scanf("%s",maze[i]);               for(j = 0 ; j < col ; ++j)               {                 if(maze[i][j] == 'S')//记录起点                 {                      start.x = i; start.y = j;                 }                 if(maze[i][j]>= 'a' && maze[i][j]<= 'e')//统计各个门所需的钥匙数量                    ++key[maze[i][j]-'a'];               }         }//for(i)         que.push(start);         if(bfs(start)) puts("YES");         else    puts("NO");         while(!que.empty())  que.pop();     }    // system("pause");     return 0;}int bfs(POS from){    POS p,next;    int i,t;    maze[from.x][from.y] = 'X';     while(!que.empty())     {         p = que.front();         que.pop();         for(i = 0 ; i < 4 ; ++i)         {            next.x = p.x + d[i][0];            next.y = p.y + d[i][1];            if(next.x>=0 && next.x < row && next.y>=0 && next.y < col)            {                  if(maze[next.x][next.y] == 'G')//找到终点，结束                   return true;                  if(maze[next.x][next.y] == '.')//可以走 进队列                  {                    que.push(next);                    maze[next.x][next.y] = 'X';                  }                  else                  {                      if(maze[next.x][next.y]>='a' && maze[next.x][next.y] <= 'e')                      {                        que.push(next);                        --key[maze[next.x][next.y]-'a'];//记录找到的钥匙数量                        maze[next.x][next.y] = '.';                      }                      else                      if(maze[next.x][next.y]>='A' && maze[next.x][next.y] <= 'E')                      {                        t = maze[next.x][next.y] - 'A';                        /*                            这个地方相当关键啊！                        */                        if(!key[t])//用钥匙把门打开即可                        {                          que.push(next);                         // maze[next.x][next.y] = 'X';                        }                        else//钥匙不够，就把原来的p进队列，先走其他的路，以寻找其他的钥匙                        {                              if(!que.empty())  que.push(p);                        }                      }                }           }//if         }//for()    }//while()    return 0;}

法二：

#include <queue>#include <stdio.h>#include <ctype.h>#include <string.h>using namespace std;typedef struct{    int x,y;}Ac;queue <Ac> S,door[5];int dir[]={0,1,0,-1,1,0,-1,0};int from_x,from_y,to_x,to_y;int lock[5],key[5];int map[25][25];bool in[25][25];bool bfs(){    while(!S.empty())    {        Ac R = S.front();        S.pop();        if(R.x == to_x && R.y == to_y)            return true;        for(int k=0;k<8;)        {            int x = R.x + dir[k++];            int y = R.y + dir[k++];            if(!in[x][y] && map[x][y])    //路或钥匙            {                S.push({x,y});                in[x][y]=1;                if(map[x][y]>0)           //钥匙                {                    key[map[x][y]-1]++;                    map[x][y]=-1;                }            }        }        if(S.empty())        {            memset(in,0,sizeof(in));            int find=0;            for(int i=0;i<5;i++)            {                if(key[i] && key[i]==lock[i])//芝麻开门                {                    key[i]=0,find=1;                    while(!door[i].empty())                    {                        R = door[i].front();                        int x = R.x;                        int y = R.y;                        map[x][y]=-1;                        door[i].pop();                    }                }            }            if(find)                S.push({from_x,from_y});            else                break;        }    }    return false;}void clear(){    while(!S.empty())        S.pop();    for(int i=0;i<5;i++)        while(!door[i].empty())            door[i].pop();}int main(){    int h,w;    while(scanf("%d%d",&h,&w),h||w)    {        getchar();        memset(lock,0,sizeof(lock));        memset(map,0,sizeof(map));        memset(key,0,sizeof(key));        memset(in,0,sizeof(in));        for(int i=1;i<=h;i++)        {            for(int j=1;j<=w;j++)            {                char c = getchar();                switch(c)                {                    case 'S':from_x=i;from_y=j;S.push({i,j});break;                    case 'G':to_x=i,to_y=j;                    case '.':map[i][j]=-1;break;                }                if(islower(c))                             //钥匙                {                    lock[c-'a']++;                    map[i][j] = c-'a'+1;                }                if(isupper(c) && c!='G' && c!='S' && c!='X')//门                {                    door[c-'A'].push({i,j});                }            }            getchar();        }        bool yes=bfs();        if(yes)            puts("YES");        else            puts("NO");        clear();    }    return 0;}

NYOJ 202

 #include <iostream>#include <cstdio>#include <string.h>using namespace std;const int N = 15;int x[N],y[N],z[N],leftp[N],rightp[N];void midorder(int x){    if(x != -1){   midorder(leftp[x]);   printf("%d\n",x);   midorder(rightp[x]);}}int main(){    int numcase;    scanf("%d",&numcase);    for(int k = 1;k <= numcase;++k)    {        int n;        scanf("%d",&n);        for(int i = 0;i < n;++i)        {           scanf("%d%d%d",&x[i],&y[i],&z[i]);        }        for(int i = 0;i < n;++i)        {            leftp[x[i]] = y[i];            rightp[x[i]] = z[i];        }        int m,xx,yy;        scanf("%d",&m);        while(m--)        {            scanf("%d%d",&xx,&yy);        }midorder(0);printf("\n");    }    return 0;}        

NYOJ 284

法一：

#include<iostream>#include<cstdio>#include<queue>using namespace std;typedef struct node{    int x,y,step;}node;bool operator<(const node &a,const node &b){    return a.step>b.step;}int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};int main(){    char map[305][305];    int i,j,ans,n,m;    node a,b;    priority_queue<node>q;    while(scanf("%d %d",&n,&m)&&(n||m))    {        a.x=0;        for(i=0;i<n;i++)        {            scanf("%s*c",&map[i]);            for(j=0;j<m;j++)            {                if(map[i][j]=='Y')                {                    a.x=i;                    a.y=j;                    a.step=0;                    q.push(a);                }            }        }        bool flag=false;        while(!q.empty())//广搜        {            a=q.top();            q.pop();            for(i=0;i<4;i++)            {                b.x=a.x+dir[i][0];                b.y=a.y+dir[i][1];                if(b.x>=0 && b.x<n && b.y>=0 && b.y<m)                {                    if(map[b.x][b.y]=='T')                    {                        ans=a.step+1;                        flag=true;                        break;                    }                    else if(map[b.x][b.y]=='E')                    {                        b.step=a.step+1;                        q.push(b);                        map[b.x][b.y]='S';                    }                    else if(map[b.x][b.y]=='B')                    {                        b.step=a.step+2;                        q.push(b);                        map[b.x][b.y]='S';                    }                }            }            if(flag)                break;        }        if(flag)            printf("%d\n",ans);        else            printf("-1\n");        while(!q.empty())        {            q.pop();        }    }    return 0;}

法二：

#include <cstdio>#include <cstring>#include <queue>using namespace std;int m, n;struct node{    int x, y;    node(int a, int b){        x = a; y = b;    }}*xm, *d;char maze[301][301];int dist[301][301], steps;bool vis[301][301];queue<node*> q;void PUSH(int a, int b){    if(vis[a][b] || a < 1 || b < 1 || a > m || b > n || maze[a][b] == 'S' || maze[a][b] == 'R') return;    q.push(new node(a, b));    dist[a][b] = steps+1;    vis[a][b] = 1;}int bfs(){    memset(vis, 0, sizeof(vis));    memset(dist, 0, sizeof(dist));    node *cur = new node(xm->x, xm->y);    while(!q.empty()) q.pop();    q.push(cur);    vis[xm->x][xm->y] = 1;    while(!q.empty()){        cur = q.front(); q.pop();        steps = dist[cur->x][cur->y];        if(cur->x == d->x && cur->y == d->y)            return dist[d->x][d->y];        if(maze[cur->x][cur->y] == 'B'){            maze[cur->x][cur->y] = 'E';            vis[cur->x][cur->y] = 0;            PUSH(cur->x, cur->y);        }else if(maze[cur->x][cur->y] != 'R' || maze[cur->x][cur->y] != 'S'){            PUSH(cur->x - 1, cur->y);            PUSH(cur->x, cur->y - 1);            PUSH(cur->x, cur->y + 1);            PUSH(cur->x + 1, cur->y);        }    }    return -1;}int main(){    int i, j;    while(scanf("%d%d", &m, &n) && m||n){        for(i = 1; i <= m; i++)            for(j = 1; j <= n; j++){                scanf(" %c ", &maze[i][j]);                if(maze[i][j] == 'Y')                    xm = new node(i, j);                else if(maze[i][j] == 'T')                    d = new node(i, j);            }        printf("%d\n", bfs());    }    return 0;}

priority_queue

#include<iostream>#include<functional>#include<queue>#include<vector>using namespace std;//定义比较结构struct cmp1{    bool operator ()(int &a,int &b){        return a>b;//最小值优先    }};struct cmp2{    bool operator ()(int &a,int &b){        return a<b;//最大值优先    }};//自定义数据结构struct number1{    int x;    bool operator < (const number1 &a) const {        return x>a.x;//最小值优先    }};struct number2{    int x;    bool operator < (const number2 &a) const {        return x<a.x;//最大值优先    }};int a[]={14,10,56,7,83,22,36,91,3,47,72,0};number1 num1[]={14,10,56,7,83,22,36,91,3,47,72,0};number2 num2[]={14,10,56,7,83,22,36,91,3,47,72,0};int main(){    priority_queue<int>que;//采用默认优先级构造队列    priority_queue<int,vector<int>,cmp1>que1;//最小值优先    priority_queue<int,vector<int>,cmp2>que2;//最大值优先    priority_queue<int,vector<int>,greater<int> >que3;//注意“>>”会被认为错误，    priority_queue<int,vector<int>,less<int> >que4;////最大值优先    priority_queue<number1>que5; //最小优先级队列    priority_queue<number2>que6;  //最大优先级队列    int i;    for(i=0;a[i];i++){        que.push(a[i]);        que1.push(a[i]);        que2.push(a[i]);        que3.push(a[i]);        que4.push(a[i]);    }    for(i=0;num1[i].x;i++)        que5.push(num1[i]);    for(i=0;num2[i].x;i++)        que6.push(num2[i]);    printf("采用默认优先关系:/n(priority_queue<int>que;)/n");    printf("Queue 0:/n");    while(!que.empty()){        printf("%3d",que.top());        que.pop();    }    puts("");    puts("");    printf("采用结构体自定义优先级方式一:/n(priority_queue<int,vector<int>,cmp>que;)/n");    printf("Queue 1:/n");    while(!que1.empty()){        printf("%3d",que1.top());        que1.pop();    }    puts("");    printf("Queue 2:/n");    while(!que2.empty()){        printf("%3d",que2.top());        que2.pop();    }    puts("");    puts("");    printf("采用头文件/"functional/"内定义优先级:/n(priority_queue<int,vector<int>,greater<int>/less<int> >que;)/n");    printf("Queue 3:/n");    while(!que3.empty()){        printf("%3d",que3.top());        que3.pop();    }    puts("");    printf("Queue 4:/n");    while(!que4.empty()){        printf("%3d",que4.top());        que4.pop();    }    puts("");    puts("");    printf("采用结构体自定义优先级方式二:/n(priority_queue<number>que)/n");    printf("Queue 5:/n");    while(!que5.empty()){        printf("%3d",que5.top());        que5.pop();    }    puts("");    printf("Queue 6:/n");    while(!que6.empty()){        printf("%3d",que6.top());        que6.pop();    }    puts("");    return 0;}/*运行结果 ：采用默认优先关系:(priority_queue<int>que;)Queue 0:83 72 56 47 36 22 14 10  7  3采用结构体自定义优先级方式一:(priority_queue<int,vector<int>,cmp>que;)Queue 1: 7 10 14 22 36 47 56 72 83 91Queue 2:83 72 56 47 36 22 14 10  7  3采用头文件"functional"内定义优先级:(priority_queue<int,vector<int>,greater<int>/less<int> >que;)Queue 3: 7 10 14 22 36 47 56 72 83 91Queue 4:83 72 56 47 36 22 14 10  7  3采用结构体自定义优先级方式二:(priority_queue<number>que)Queue 5: 7 10 14 22 36 47 56 72 83 91Queue 6:83 72 56 47 36 22 14 10  7  3*/

法三：

#include <iostream> #include <queue> #include <cstdio> #include <cstring> using namespace std; #define  max 301 char map[max][max]; int n,m; bool vist[max][max]; int d[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; struct node {     int x,y,c;     friend bool operator <(const node &s1,const node &s2)     {         return s1.c>s2.c;     } }; node s,e; int bfs() {     priority_queue < node > q;     memset(vist,0,sizeof(vist));     node now,t;     s.c=0;     q.push(s);     vist[s.x][s.y]=1;     while (!q.empty())     {         now=q.top();         q.pop();         if (now.x==e.x&&now.y==e.y)         {             return now.c;         }         for (int i=0;i<4;i++)         {             t.x=now.x+d[i][0];             t.y=now.y+d[i][1];             if (t.x>=0&&t.x<n&&t.y>=0&&t.y<m&&!vist[t.x][t.y])             {                 vist[t.x][t.y]=1;                 if (map[t.x][t.y]=='B')                 {                     t.c=now.c+2;                     q.push(t);                 }                 else                     if (map[t.x][t.y]=='E'||map[t.x][t.y]=='T')                     {                         t.c=now.c+1;                         q.push(t);                     }             }         }     }     return -1; } int main() {     int i,j;     while (cin>>n>>m,m+n)     {         for (i=0;i<n;i++)         {             for (j=0;j<m;j++)             {                 cin>>map[i][j];                 if (map[i][j]=='Y')                     s.x=i,s.y=j;                 if (map[i][j]=='T')                     e.x=i,e.y=j;             }         }         int sum=bfs();      cout<<sum<<endl;     }     return 0; }

NYOJ 325

 #include <stdio.h>#define max(a,b) a>b?a:bint V,ans,n,w[21],sum[21];void dfs(int i,int cnt){    if(i == 0)    {        ans = max(ans,cnt);        return ;    }    if(ans == V || cnt+sum[i] <= ans)       //cut        return ;    if(cnt+w[i] <= V)        dfs(i-1,cnt+w[i]);    dfs(i-1,cnt);}int main(){    while(~scanf("%d",&n))    {        ans = 0;        for(int i=1;i<=n;i++)        {            scanf("%d",&w[i]);            sum[i] = sum[i-1] + w[i];        }        V = sum[n]/2;        dfs(n,0);        printf("%d\n",sum[n]-2*ans);    }    return 0;}        

NYOJ 353

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int MAXN = 35;const int INF = 0xfffffff;struct Node{    int x, y, z;    int step;    Node()    {        x = y = z = 0;        step = 0;    }};Node TargetPos;char Graph[MAXN][MAXN][MAXN];int MinTime[MAXN][MAXN][MAXN];int row, col, level;int dir[6][3] = {{1, 0, 0}, {0, 1, 0}, {-1, 0, 0}, {0, -1, 0}, {0, 0, 1}, {0, 0, -1}};bool Is_CanGo(Node CurPos){    if(CurPos.x < 0 || CurPos.x >= level || CurPos.y < 0 || CurPos.y >= row || CurPos.z < 0 || CurPos.z >= col || Graph[CurPos.x][CurPos.y][CurPos.z] == '#')        return false;    return true;}void BFS(Node S){    queue <Node> Que;    Que.push(S);    while(!Que.empty())    {        Node Curpos = Que.front();        Que.pop();        for(int i = 0; i < 6; ++i)        {            Node t;            t.x = Curpos.x + dir[i][0];            t.y = Curpos.y + dir[i][1];            t.z = Curpos.z + dir[i][2];            if(Is_CanGo(t))            {                t.step = Curpos.step + 1;                if(t.step < MinTime[t.x][t.y][t.z])                {                    MinTime[t.x][t.y][t.z] = t.step;                    Que.push(t);                }            }        }    }}int main(){    while(scanf("%d %d %d", &level, &row, &col) && (level + row + col))    {        Node StartPos;        for(int i = 0; i < level; ++i)        {            for(int j = 0; j < row; ++j)            {                scanf("%s", Graph[i][j]);                for(int z = 0; z < col; ++z)                {                    MinTime[i][j][z] = INF;                    if(Graph[i][j][z] == 'S')                        StartPos.x = i, StartPos.y = j, StartPos.z = z;                    else if(Graph[i][j][z] == 'E')                        TargetPos.x = i, TargetPos.y = j, TargetPos.z = z;                }                getchar();            }           if(i != level - 1)                getchar();        }        StartPos.step = 0;        BFS( StartPos );        int t;        t = MinTime[TargetPos.x][TargetPos.y][TargetPos.z];        if(t != INF)            printf("Escaped in %d minute(s).\n", t);        else            printf("Trapped!\n");    }    return 0;}

NYOJ 488

#include<stdio.h>#include<string.h>int Map[9][9],floag;int IsCan(int number,int x,int y){//判断Numebr是否满足数独的条件    int i,j;    for(i=0;i<9;i++){        if(Map[i][y]==number){            return 0;        }    }    for(i=0;i<9;i++){        if(Map[x][i]==number){            return 0;        }    }    x=x/3*3,y=y/3*3;    for(i=x;i<x+3;i++)        for(j=y;j<y+3;j++)            if(Map[i][j]==number)            return 0;    return 1;}void dfs(int x,int y){    if(floag) return ;//floag用于标记，当所有数据都搜完的时候就给跳出    if(x==9 && y==0){        int i,j;        for(i=0;i<9;i++){            for(j=0;j<9;j++){                printf("%d ",Map[i][j]);            }                printf("\n");        }            floag=1;        return ;    }    if(y==9){//一行结束进入下一行        dfs(x+1,0);    }    if(Map[x][y]!=0){//已经有数就填下一个        dfs(x,y+1);    }    if(Map[x][y]==0){        int i;        for(i=1;i<=9;i++){            if(IsCan(i,x,y)){                Map[x][y]=i;                dfs(x,y+1);                Map[x][y]=0;//回溯清除，当一次搜索结束的时候但是不满足就将之前的结果清除            }        }    }}int main(){    int n;    scanf("%d",&n);    while(n--){        int i,j;        for(i=0;i<9;i++){            for(j=0;j<9;j++){                scanf("%d",&Map[i][j]);            }        }        dfs(0,0);        memset(Map,0,sizeof(Map));        floag=0;    }    return 0;}

NYOJ 491

法一：

#include"iostream"#include<cstring>#include<stdio.h>#include<time.h>using namespace std;typedef unsigned char uchar;//char cc[2]={'+','-'};    //便于输出int n,                     //第一行符号总数    half,                 //全部符号总数一半    counter;             //1计数，即  '-' 号计数char **p;                //符号存储空间long sum;                //符合条件的三角形计数//t，第一行第 t个符号void Backtrace(int t){    int i, j;    if( t > n )        sum++;    else    {       for(i=0; i<2; ++i)  //只取  0('+')  或者  1('-')       {            p[1][t] = i;   //第一行第 t个符号            counter += i;        //'-'号统计            for(j=2; j<=t; ++j)    //当第一行符号 >=2时，可以运算出下面行的某些符号（第一行有几个数就可以相应往下计算几行，每次计算过的就不用重复计算）            {                p[j][t-j+1] = p[j-1][t-j+1]^p[j-1][t-j+2];//通过异或运算下行符号                counter += p[j][t-j+1];            }            if( (counter <= half) && ( t*(t+1)/2 - counter <= half) )//若符号统计未超过半数，并且另一种符号也未超过半数                Backtrace(t+1);            //在第一行增加下一个符号            //回溯，判断另一种符号情况            for(j=2; j<=t; ++j)                counter -= p[j][t-j+1];            counter -= i;       }    }}int main(){    while(scanf("%d", &n) != EOF)    {        counter = 0;        sum = 0;        half = n*(n+1)/2;        if( half%2 == 0 )//总数须为偶数，若为奇数则无解        {            half /= 2;            p = new char *[n+1];            for(int i=0; i<=n; ++i)            {               p[i] = new char[n+1];               memset(p[i], 0, sizeof(char)*(n+1));            }            Backtrace(1);        }        printf("%d\n", sum);    }    return 0;}

法二：

#include<iostream>#include<cstring>#include<cstdio>#include<ctime>#include<algorithm>using namespace std;#define M 20int cnt,zcnt,fcnt,res[M];bool ope[M][M];void dfs1(int n,int p){if(p == n){zcnt = fcnt = 0;for(int i = 1;i <= p;++i)//for(int j = 1;j <= p - i + 1;++j){if(ope[i-1][j]) fcnt++;else zcnt++;if(j == p - i + 1) continue;if(ope[i-1][j] == ope[i-1][j+1]) ope[i][j] = false;else ope[i][j] = true;}if(zcnt == fcnt && zcnt) cnt++;return ;}dfs1(n,p+1);ope[0][p+1] = true;dfs1(n,p+1);ope[0][p+1] = false;}void init(){for(int i = 1;i < 20;++i){memset(ope,0,sizeof(ope));cnt = 0;if((i*(i+1)/2)%2==0) dfs1(i,0);res[i] = cnt;}}int main(){int n;init();while(~scanf("%d",&n)){printf("%d\n",res[n]);}return 0;}/*枚举第一行所有的情况，可以用深搜实现，可根据第一行逐行确定出下一行的状态，，然后算出每种情况两种符号的个数，在运算之前可以先进行一下判断，当(n+1)*n/2为奇数时，结果为零*///超时

NYOJ 499

法一：

#include<stdio.h>#include<string.h>int a[10][10], count, n, m;int b[10][10];void DFS(int x, int y){if(x > n || x < 1 || y > m || y < 1 || a[x][y] == 1){return ;}if(x == n && y == m){count++;return ;}printf("%d,%d\n",x,y);a[x][y] = 1;DFS(x + 1, y);DFS(x - 1, y);DFS(x, y + 1);DFS(x, y - 1);a[x][y] = 0;}int main(){int t, i, j;scanf("%d", &t);while(t--){count = 0;memset(b, 0, sizeof(b));scanf("%d%d", &n, &m);for(i = 1; i <= n; i++){for(j = 1; j <= m; j++){scanf("%d", &a[i][j]);}}DFS(1, 1);printf("%d\n", count);}return 0;}/*0 1 1 1 1 1 1 10 0 0 0 0 0 0 10 0 0 0 0 0 0 10 0 0 0 0 0 0 10 0 0 0 0 0 0 10 0 0 0 0 0 0 11 1 1 1 1 1 1 1*/

法二：

#include <stdio.h>int count;int dt[8][8];int r,c;void init(){int i;for(i = 0 ; i <= r+1; i++){dt[i][0] = 1;dt[i][c+1] = 1;}for(i = 0; i <= c+1; i++){dt[0][i] = 1;dt[r+1][i] = 1;}}void print(){int i,j;for(i = 0; i <= r+1; i++){for(j = 0; j <= c+1; j++)printf("%d ",dt[i][j]);printf("\n");}}void dfs(int x,int y){if(x == r && y == c){count++;return;}dt[x][y] = 1;if(!dt[x+1][y]) dfs(x+1,y);if(!dt[x][y+1]) dfs(x,y+1);if(!dt[x-1][y]) dfs(x-1,y);if(!dt[x][y-1]) dfs(x,y-1);dt[x][y] = 0;}int main(){int i,j,t;scanf("%d",&t);while(t--){scanf("%d%d",&r,&c);count = 0;init();for(i = 0; i < r; i++)for(j = 0; j < c; j++)scanf("%d", &dt[i+1][j+1]);dfs(1,1);printf("%d\n",count);}return 0;}

NYOJ 523

法一：

#include <stdio.h>#include <string.h>#include <queue>using namespace std;int map[55][55][55],f[]={1,-1,0,0,0,0},g[]={0,0,1,-1,0,0},w[]={0,0,0,0,1,-1};bool vis[55][55][55],flag;struct Node{    int x,y,z,step;};int bfs(int a,int b,int c,int time){    int i,j,k;     int r,s,t;    memset(vis,0,sizeof(vis));    queue <Node> q;    Node temp={0,0,0,time};    vis[0][0][0]=1;    q.push(temp);    while(!q.empty())    {        temp=q.front();        q.pop();        for(i=0;i<6;i++)        {            r=temp.x+f[i];            s=temp.y+g[i];            t=temp.z+w[i];            if(r>=0&&r<a&&s>=0&&s<b&&t>=0&&t<c&&vis[r][s][t]==0)            {                if(r==a-1&&s==b-1&&t==c-1&&map[r][s][t]==0&&temp.step-1>=0)                {                    flag=1;                    break;                }                else if(map[r][s][t]==0&&temp.step-1)                {                    Node next={r,s,t,temp.step-1};                    q.push(next);                    vis[r][s][t]=1;                }            }        }        if(flag==1||temp.step-1==0)            break;    }    while(!q.empty())        q.pop();    return time-(temp.step-1);}int main(){    int a,b,c;     int i,j,k,T;     int time;    scanf("%d",&T);    while(T--)    {        memset(map,0,sizeof(map));        flag=0;        scanf("%d%d%d%d",&a,&b,&c,&time);        for(k=0;k<a;k++)        {            for(i=0;i<b;i++)            {                for(j=0;j<c;j++)                {                    scanf("%d",&map[k][i][j]);                }            }        }        int cnt=bfs(a,b,c,time);        if(flag==1)            printf("%d\n",cnt);        else            printf("-1\n");    }    return 0;}

法二：

#include<iostream>#include<cstring>#include<cstdio>#include<queue>#include<algorithm>using namespace std;const int N = 55;const int dir[6][3] = {{0, 0, 1}, {0, 0, -1}, {-1, 0, 0}, {1, 0, 0}, {0, 1, 0}, {0, -1, 0}};struct node{int x, y, z;int ceng;}que[50000];int maze[N][N][N];int a, b, c, t;int num;int head, tail;int Scan(){int res = 0 , ch;while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) ){if( ch == EOF )  return 1 << 30 ;}res = ch - '0' ;while( ( ch = getchar() ) >= '0' && ch <= '9' )res = res * 10 + ( ch - '0' ) ;return res ;}int BFS(){node cur;cur.x = 1, cur.y = 1, cur.z = 1, cur.ceng = 0;maze[cur.x][cur.y][cur.z] = 0;que[tail++] = cur;while(head < tail){node temp = que[head++];if(temp.ceng > t) //无法到达return -1;if(temp.x == a && temp.y == b && temp.z == c && temp.ceng <= t)return temp.ceng;for(int i = 0; i < 6; ++i){node now;now.x = temp.x + dir[i][0]; now.y = temp.y + dir[i][1]; now.z = temp.z + dir[i][2];if(maze[now.x][now.y][now.z]){now.ceng = temp.ceng + 1;que[tail++] = now;maze[now.x][now.y][now.z] = 0;}}}return -1;}int main(){int ncase;ncase = Scan();while(ncase--){num = head = tail = 0;memset(maze, 0, sizeof(maze));a = Scan(), b = Scan(), c = Scan(), t = Scan();for(int i = 1; i <= a; ++i)for(int j = 1; j <= b; ++j)for(int k = 1; k <= c; ++k){maze[i][j][k] = Scan() ^ 1;num += maze[i][j][k];}if(maze[a][b][c] == 0 || num < a + b + c - 3) //终点无法走{printf("-1\n");continue;}printf("%d\n", BFS());}return 0;}

NYOJ 592

#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>using namespace std;int map[105][105]={0};int mm=10000,n=100;int fir,end,loop[105][105];int prime[11000]={1,1,0};int sx[]={0,0,1,-1},zy[]={1,-1,0,0};void Build_map() //建立地图{    int i,j,k;    for(i=1;i<=(n+1)/2;i++)    {        for(j=1;j<=n-2*i+2;j++)            map[i][j+i-1]=mm--;        for(j=1;j<=n-2*i;j++)           map[j+i][n-i+1]=mm--;        for(j=n-i+1;j>=i;j--)           map[n-i+1][j]=mm--;        for(j=n-i;j>=i+1;j--)           map[j][i]=mm--;    }}void printf_map()//打印地图{    int i,j;    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)          printf("%d ",map[i][j]);        printf("\n");    }}void judge_prime()  //筛选法打个素数表{    int i,j;    for(i=2;i<=10000;i++)    {        if(prime[i]==0)        {            for(j=i+i;j<=10000;j+=i)              prime[j]=1;        }    }}struct coordinate{    int x;    int y;    int step;}t1;void find() //找到 fir 的坐标，存在 t1 中{    int i,j;    for(i=1;i<=n;i++)      for(j=1;j<=n;j++)      {          if(map[i][j]==fir)          {              t1.x=i;              t1.y=j;              loop[i][j]=1;              t1.step=0;              return;          }      }}int bfs(){    int i,j,step,x,y;    queue<coordinate> Q;    Q.push(t1);    while(!Q.empty())    {        i=Q.front().x;        j=Q.front().y;        step=Q.front().step;        Q.pop();        if(map[i][j]==end)          return step;        for(int a=0;a<4;a++)        {            x=i+sx[a];            y=j+zy[a];            if(map[x][y]!=0&&prime[map[x][y]]==1&&loop[x][y]==0)            {                coordinate t2={x,y,step+1};                Q.push(t2);                loop[x][y]=1;            }        }    }    return 0;}int main(){    int i,j,nn=0;    Build_map();    judge_prime();    //printf_map();    while(~scanf("%d%d",&fir,&end))    {        nn++;        memset(loop,0,sizeof(loop));        find();        int ans=bfs();        if(ans)            printf("Case %d: %d\n",nn,ans);        else            printf("Case %d: impossible\n",nn);    }}

NYOJ 722

#include<stdio.h>#include<string.h>int Map[9][9],floag;int IsCan(int number,int x,int y){//判断Numebr是否满足数独的条件    int i,j;    for(i=0;i<9;i++){        if(Map[i][y]==number){            return 0;        }    }    for(i=0;i<9;i++){        if(Map[x][i]==number){            return 0;        }    }    x=x/3*3,y=y/3*3;    for(i=x;i<x+3;i++)        for(j=y;j<y+3;j++)            if(Map[i][j]==number)            return 0;    return 1;}void dfs(int x,int y){    if(floag) return ;//floag用于标记，当所有数据都搜完的时候就给跳出    if(x==9 && y==0){        int i,j;        for(i=0;i<9;i++){            for(j=0;j<9;j++){                printf("%d ",Map[i][j]);            }                printf("\n");        }            floag=1;        return ;    }    if(y==9){//一行结束进入下一行        dfs(x+1,0);    }    if(Map[x][y]!=0){//已经有数就填下一个        dfs(x,y+1);    }    if(Map[x][y]==0){        int i;        for(i=1;i<=9;i++){            if(IsCan(i,x,y)){                Map[x][y]=i;                dfs(x,y+1);                Map[x][y]=0;//回溯清除，当一次搜索结束的时候但是不满足就将之前的结果清除            }        }    }}int main(){    int n;    scanf("%d",&n);    while(n--){        int i,j;        for(i=0;i<9;i++){            for(j=0;j<9;j++){                scanf("%d",&Map[i][j]);            }        }        dfs(0,0);        memset(Map,0,sizeof(Map));        floag=0;    }    return 0;}