poj2251 Dungeon Master 简单BFS

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题目链接：http://poj.org/problem?id=2251

题意：从S走到E，若走通，输出Escaped in X minute(s).若走不出，输出"Trapped!"

方法：简单BFS;

#include<cstdio>#include<iostream>#include<cstring>#include<queue>using namespace std;char dun[40][40][40];bool vis[40][40][40];int L,R,C,fx[6][3]={{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};struct node{int l,r,c,tim;}ic,tmp,s,e;void bfs(){memset(vis,0,sizeof(vis));queue<node>q;q.push(s);vis[s.l][s.r][s.c]=1;while(!q.empty()){ic=q.front();q.pop();if(ic.l==e.l&&ic.r==e.r&&ic.c==e.c){e.tim=ic.tim;return ;} for(int i=0;i<6;++i){int l,r,c;l=ic.l+fx[i][0];r=ic.r+fx[i][1];c=ic.c+fx[i][2];if(l<0||r<0||c<0||l>=L||r>=R||c>=C)continue;if(vis[l][r][c]||dun[l][r][c]=='#')continue;vis[l][r][c]=1;tmp.l=l;tmp.r=r;tmp.c=c;tmp.tim=ic.tim+1;q.push(tmp);}}}int main(){while(scanf("%d%d%d",&L,&R,&C)&&(L||R||C)){for(int i=0;i<L;++i)    for(int j=0;j<R;++j)        for(int k=0;k<C;++k)            {                cin>>dun[i][j][k];                if(dun[i][j][k]=='S')                {                s.l=i;                s.r=j;                s.c=k;                s.tim=0;}if(dun[i][j][k]=='E'){e.l=i;e.r=j;e.c=k;e.tim=0;}}bfs();if(!e.tim)cout<<"Trapped!"<<endl;else cout<<"Escaped in "<<e.tim<<" minute(s)."<<endl;}}

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