基础BFS 邻接矩阵Flood fill 算法题目总结

很基础的一类BFS题，个人觉得BFS在求最短、最近、最快等题目时比DFS还是有较大优势的

Poj 1979 Red and Black

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int N=25;struct Point{int x,y;};int graph[N][N];bool visit[N][N];int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};int W,H,u,v;int OK (int x,int y){return x>=0 && x<W && y>=0 && y<H;}int BFS (){int ans=1;Point top,t;    queue <Point> Q;top.x=u;top.y=v;Q.push(top);visit[top.x][top.y]=true;while (Q.empty()==false){top=Q.front();Q.pop();for (int i=0;i<4;i++){t.x=top.x+dx[i];t.y=top.y+dy[i];if (OK(t.x,t.y) && graph[t.x][t.y] && !visit[t.x][t.y]){visit[t.x][t.y]=true;Q.push(t);ans++;}}}return ans;}int main (){while (scanf("%d%d",&W,&H) , (W || H)){        char str[N];        memset(visit,false,sizeof(visit));        memset(graph,0,sizeof(graph));        for (int i=0;i<H;i++)        {            scanf("%s",str);            for (int j=0;j<W;j++)            {                if (str[j] == '.')                    graph[j][i]=1;                if (str[j] == '@')                {                    graph[j][i]=1;   //u,v存起点坐标                    u=j;                    v=i;                }            }        }        printf("%d\n",BFS());    }return 0;}

Poj 2243 Knight Moves

题意：要求用象棋中马跳动方式到达终点，求马需要跳动的最少次数。

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int N=10;struct Point{int x,y,step;};int X,Y,n,u,v;bool visit[N][N];int dx[8] = {1,2,2,1,-1,-2,-2,-1};int dy[8] = {2,1,-1,-2,-2,-1,1,2};int OK (int x,int y){return x>=0 && x<=7 && y>=0 && y<=7 && visit[x][y]==false;}int BFS (){Point top,t;queue <Point> Q;top.step=0;top.x=u;top.y=v;Q.push(top);visit[top.x][top.y]=true;while (Q.empty()==false){top=Q.front();Q.pop();if (top.x==X && top.y==Y)break;for (int i=0;i<8;i++){t.x=top.x+dx[i];t.y=top.y+dy[i];t.step=top.step+1;if (OK(t.x,t.y) && !visit[t.x][t.y]){visit[t.x][t.y]=true;Q.push(t);}}}return top.step;}int main (){char ch1[4],ch2[4];while (~scanf("%s%s",&ch1,&ch2)){memset(visit,false,sizeof(visit));u=ch1[0]-'a';v=ch1[1]-'1';X=ch2[0]-'a';Y=ch2[1]-'1';printf("To get from %s to %s takes %d knight moves.\n",ch1,ch2,BFS());}return 0;}

Poj 1915 Knight Moves

题意：输入n代表有个n*n的棋盘，输入开始位置的坐标和结束位置的坐标，
问一个骑士朝棋盘的八个方向走马字步，从开始坐标到结束坐标可以经过多少步。

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int N=305;struct Point{int x,y,step;};int X,Y,n,u,v;bool visit[N][N];int dx[8] = {1,2,2,1,-1,-2,-2,-1};int dy[8] = {2,1,-1,-2,-2,-1,1,2};queue <Point> Q;int OK (int x,int y){return x>=0 && x<=n-1 && y>=0 && y<=n-1 && visit[x][y]==false;}int BFS (){Point top,t;int i;top.step=0;top.x=u;top.y=v;Q.push(top);visit[top.x][top.y]=true;while (Q.empty()==false){top=Q.front();Q.pop();if (top.x==X && top.y==Y)break;for (i=0;i<8;i++){t.x=top.x+dx[i];t.y=top.y+dy[i];t.step=top.step+1;if (OK(t.x,t.y)){visit[t.x][t.y]=true;Q.push(t);}}}return top.step;}int main (){int T;scanf("%d",&T);while (T--){memset(visit,false,sizeof(visit));while (Q.empty()==false)            Q.pop();scanf("%d%d%d%d%d",&n,&X,&Y,&u,&v);printf("%d\n",BFS());}return 0;}

LightOJ 1012 Guilty Prince

题意:给两个数,w,h,表示给一个 高h,宽w的矩阵,
‘#’表示不能走,‘.’表示能走,‘@’表示起始点,求从起始点出发能访问多少个点。

#include <queue>#include <cstdio>#include <cstring>using namespace std;const int N=25;char g[N][N];bool visit[N][N];int w,h;int dx[4] = {1,-1,0,0};int dy[4] = {0,0,1,-1};struct Node{int x,y;}temp,cur,s;bool OK (Node a){if (visit[a.x][a.y]==false && g[a.x][a.y]!='#')return true;return false;}int BFS (){int cnt=0;queue<Node> q;q.push(s);while (!q.empty()){cur = q.front();q.pop();for (int i=0;i<4;i++){temp.x=cur.x+dx[i];temp.y=cur.y+dy[i];if (OK(temp)){visit[temp.x][temp.y]=true;q.push(temp);cnt++;}}}return cnt;}int main (){int T;scanf("%d",&T);for (int Cas=1;Cas<=T;Cas++){scanf("%d%d",&w,&h);memset(g,'#',sizeof(g));memset(visit,false,sizeof(visit));for (int i=1;i<=h;i++){scanf("%s",&g[i][1]);g[i][w+1]='#';for (int j=1;j<=w;j++)if (g[i][j]=='@'){s.x=i;s.y=j;visit[s.x][s.y]=true;}}printf("Case %d: %d\n",Cas,BFS()+1);}return 0;}

Poj 3626 Mud Puddles

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int N=1005;struct Point{int x,y,step;};int a[N][N],X,Y,n;bool visit[N][N];int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};int OK (int x,int y){return x>=0 && x<=1000 && y>=0 && y<=1000 && visit[x][y]==false;}int BFS (){Point top,t;queue <Point> Q;top.step=0;top.x=top.y=500;Q.push(top);visit[top.x][top.y]=true;while (Q.empty()==false){top=Q.front();Q.pop();if (top.x==X && top.y==Y)break;for (int i=0;i<4;i++){t.x=top.x+dx[i];t.y=top.y+dy[i];t.step=top.step+1;if (OK(t.x,t.y)){visit[t.x][t.y]=true;Q.push(t);}}}return top.step;}int main (){while (~scanf("%d%d%d",&X,&Y,&n)){int i,x,y;memset(visit,false,sizeof(visit));for (i=1;i<=n;i++)scanf("%d%d",&x,&y),visit[x+500][y+500]=true;X+=500,Y+=500;printf("%d\n",BFS());}return 0;}

Poj 2251 Dungeon Master

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int N=50;struct Point{int x,y,z,step;};int visit[N][N][N],X,Y,Z,x,y,z;int L,R,C;int dx[]={1,-1,0,0,0,0};int dy[]={0,0,1,-1,0,0};int dz[]={0,0,0,0,1,-1};int OK (int x,int y,int z){return x>=1 && x<=R && y>=1 && y<=C && z>=1 && z<=L;}void Input (){char str[N];memset(visit,true,sizeof(visit));for (int k=1;k<=L;k++) for (int i=1;i<=R;i++)    {        scanf("%s",str+1);        for (int j=1;j<=C;j++)        {            if (str[j]=='.')                visit[i][j][k]=false;            else if(str[j]=='S')x=i,y=j,z=k;            else if(str[j]=='E')            {                X=i,Y=j,Z=k;                visit[i][j][k]=false;            }        }    }}int BFS (){Point top,t;queue <Point> Q;int time=9999;top.step=0;top.x=x;top.y=y;top.z=z;Q.push(top);visit[top.x][top.y][top.z]=true;while (Q.empty()==false){top=Q.front();Q.pop();if (top.x==X && top.y==Y && top.z==Z){time=top.step;break;}for (int i=0;i<6;i++){t.x=top.x+dx[i];t.y=top.y+dy[i];t.z=top.z+dz[i];t.step=top.step+1;if (OK(t.x,t.y,t.z) && visit[t.x][t.y][t.z]==false){visit[t.x][t.y][t.z]=true;Q.push(t);}}}return time;}int main (){while (scanf("%d%d%d",&L,&R,&C) && (L||R||C)){Input();int ans=BFS();if (ans==9999)printf("Trapped!\n");elseprintf("Escaped in %d minute(s).\n",ans);}return 0;}