小试牛刀-搜索基础入门（杭电五题）

 hdu 1241 Oil Deposits

水题，BFS，判断区域的块数。

代码清单：

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef pair<int,int>P;int m,n;char s[105][105];int xy[8][2]={{0,-1},{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1}};void bfs(int x,int y){    queue<P>q;    s[x][y]='*';    q.push(P(x,y));    while(q.size()){        P p=q.front(); q.pop();        for(int i=0;i<8;i++){            int xx=p.first+xy[i][0];            int yy=p.second+xy[i][1];            if(xx>=0&&xx<m&&yy>=0&&yy<n&&s[xx][yy]!='*'){                s[xx][yy]='*';                q.push(P(xx,yy));            }        }    }}int main(){    while(scanf("%d%d",&m,&n)!=EOF){        if(m==0&&n==0) break;        //memset(s,'\0',sizeof(s));        for(int i=0;i<m;i++)            scanf("%s",s[i]);        int ans=0;        for(int i=0;i<m;i++){            for(int j=0;j<n;j++){                if(s[i][j]=='@'){                    ans++;                    bfs(i,j);                }            }        }        printf("%d\n",ans);    }return 0;}

hdu1312 Red and Black

水题，BFS（DFS），求可走点的总和。

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef pair<int,int>P;int m,n;int sx,sy;int vis[25][25];char s[25][25];int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};int bfs(){    int sum=0;    queue<P>q;    while(q.size()) q.pop();    q.push(P(sx,sy));    vis[sx][sy]=1;    while(q.size()){        P p=q.front();        q.pop();        sum++;        for(int i=0;i<4;i++){            int xx=p.first+xy[i][0];            int yy=p.second+xy[i][1];            if(xx>=0&&xx<n&&yy>=0&&yy<m&&s[xx][yy]!='#'&&!vis[xx][yy]){                vis[xx][yy]=1;  //cout<<xx<<" "<<yy<<endl;                q.push(P(xx,yy));            }        }    }return sum;}int main(){    while(scanf("%d%d",&m,&n)!=EOF){        if(m==0&&n==0) break;        for(int i=0;i<n;i++){            scanf("%s",s[i]);            for(int j=0;j<m;j++){                if(s[i][j]=='@'){                    sx=i;                    sy=j;                }            }        }        memset(vis,0,sizeof(vis));        printf("%d\n",bfs());    }return 0;}

hdu 1010 Tempter of the Bone

DFS，从中学到了奇偶性剪枝。

题意：迷宫中只有一扇门，且只在第K秒时开门，问能否走出迷宫。

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int first;int N,M,T;int sx,sy;int ex,ey;char s[8][8];int x[4]={1,-1,0,0};int y[4]={0,0,-1,1};void dfs(int X,int Y,int t){    if(t==T){        if(X==ex&&Y==ey)            first=1;        return ;    }    if(first) return ;    int judge=abs(X-ex)+abs(Y-ey)-abs(t-T);  //奇偶性剪枝:judge<=0且为偶数才可以继续    if(judge>0 || judge%2) return ;    for(int i=0;i<4;i++){        int xx=X+x[i];        int yy=Y+y[i];        if(xx>=0&&xx<N&&yy>=0&&yy<M&&s[xx][yy]!='X'){            s[xx][yy]='X';            dfs(xx,yy,t+1);            s[xx][yy]='.';        }    }}int main(){    while(scanf("%d%d%d",&N,&M,&T)!=EOF){        if(N==0&&M==0&&T==0) break;        int pos=0;        for(int i=0;i<N;i++){            scanf("%s",s[i]);            for(int j=0;j<M;j++){                if(s[i][j]=='S'){                    sx=i;                    sy=j;                }                else if(s[i][j]=='D'){                    ex=i;                    ey=j;                }                else if(s[i][j]=='X'){                    pos++;                }            }        }        if(N*M-pos<=T) printf("NO\n"); //可走的点数必须大于T        else{            first=0;            s[sx][sy]='X';            dfs(sx,sy,0);            if(first) printf("YES\n");            else      printf("NO\n");        }    }return 0;}

 hdu 1242 Rescue

优先队列+BFS。注意救援有多个，所以以终点为起始点，只要走到一个最近的救援点即可。

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct edge{    int x,y;    int t;    friend bool operator<(edge a,edge b){        return a.t>b.t;    }};int N,M;int ex,ey,ok;int vis[205][205];char s[205][205];int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};int bfs(){    priority_queue<edge>q;    while(q.size()) q.pop();    edge p;    p.x=ex; p.y=ey; p.t=0;    vis[ex][ey]=1;    q.push(p);    while(q.size()){        p=q.top();        q.pop();        if(s[p.x][p.y]=='r'){            return p.t;        }        for(int i=0;i<4;i++){            edge w;            w.x=p.x+xy[i][0];            w.y=p.y+xy[i][1];            if(w.x>=0&&w.x<N&&w.y>=0&&w.y<M&&s[w.x][w.y]!='#'&&!vis[w.x][w.y]){                w.t=p.t+1;                if(s[w.x][w.y]=='x') w.t++;                vis[w.x][w.y]=1;                q.push(w);            }        }    }    return -1;}int main(){    while(scanf("%d%d",&N,&M)!=EOF){        for(int i=0;i<N;i++){            scanf("%s",s[i]);            for(int j=0;j<M;j++){                if(s[i][j]=='a'){                    ex=i;                    ey=j;                }            }        }        memset(vis,0,sizeof(vis));        int ans=bfs();        if(ans!=-1) printf("%d\n",ans);        else        printf("Poor ANGEL has to stay in the prison all his life.\n");    }return 0;}

hdu 1026  Ignatius and the Princess I

优先队列+BFS+路径还原。

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct edge{    int x,y;    int t,r;    friend bool operator<(edge c,edge d){        return c.t>d.t;    }}a[10000],b[105][105];  //b数组记录路径int N,M;int vis[105][105];char s[105][105];int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};void bfs(){    priority_queue<edge>q;    while(q.size()) q.pop();    edge p,w;    p.x=0; p.y=0; p.t=0,p.r=0;    vis[0][0]=1;    q.push(p);    int first=0;    while(q.size())    {        p=q.top(); q.pop();        if(p.x==N-1&&p.y==M-1){            first=1;            break;        }        for(int i=0;i<4;i++){            w.x=p.x+xy[i][0];            w.y=p.y+xy[i][1];            if(w.x>=0&&w.x<N&&w.y>=0&&w.y<M&&s[w.x][w.y]!='X'&&!vis[w.x][w.y]){                b[w.x][w.y].x=p.x;                b[w.x][w.y].y=p.y;                w.t=p.t+1;                w.r=p.r+1;                vis[w.x][w.y]=1;                if(s[w.x][w.y]!='.')                    w.t=w.t+s[w.x][w.y]-'0';                q.push(w);            }        }    }    if(first)    {        int k=1;        int i=p.r-1,xx=p.x,yy=p.y;        a[p.r].x=p.x; a[p.r].y=p.y;        for(i=p.r-1;i>=0;i--){            a[i].x=b[xx][yy].x;            a[i].y=b[xx][yy].y;            xx=a[i].x;            yy=a[i].y;        }        printf("It takes %d seconds to reach the target position, let me show you the way.\n",p.t);        for(int i=0;i<p.r;i++){            printf("%ds:(%d,%d)->(%d,%d)\n",k++,a[i].x,a[i].y,a[i+1].x,a[i+1].y);            if(s[a[i+1].x][a[i+1].y]!='.'&&s[a[i+1].x][a[i+1].y]!='X'){                int tt=s[a[i+1].x][a[i+1].y]-'0';                while(tt--) printf("%ds:FIGHT AT (%d,%d)\n",k++,a[i+1].x,a[i+1].y);            }        }    }    else printf("God please help our poor hero.\n");    printf("FINISH\n");}int main(){    while(scanf("%d%d",&N,&M)!=EOF){        for(int i=0;i<N;i++)            scanf("%s",s[i]);        memset(vis,0,sizeof(vis));        bfs();    }return 0;}