UVA 11624-Fire!-BFS-

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http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=28833

题意：就是找起点S到走出n*m格子外的最短时间

开始有一些Fire点，每单位时间会四处延伸

先预处理好每个格子最先着火的时间。

在bfs找最短路的基础上加多一个判断是否着火即可

复杂度（NM）

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <iostream>using namespace std;const double pi=acos(-1.0);double eps=0.000001; int min(int a,int b){return a<b?a:b;} int max(int a,int b){return a>b?a:b;}int r,c;int stx,sty;struct node{int x,y,t;node(){}node(int a,int b,int c){x=a,y=b;t=c;}}; int ok=0;char tm[1005][1005];int vis[1005][1005];//着火时间int dx[]={0,0,1,-1};int dy[]={1,-1,0,0};int visited[1005][1005];//访问过queue <node> q,que; void fire(){while(!que.empty()){//printf("%d\n",que.size());node tt=que.front();que.pop();int x=tt.x;int y=tt.y;int cur=tt.t;for (int i=0;i<4;i++){int xx=x+dx[i];int yy=y+dy[i];if (!(xx>=1&&xx<=r&&yy>=1&&yy<=c))continue;if (tm[xx][yy]=='#')continue;if (vis[xx][yy]!=-1) continue;vis[xx][yy]=cur+1;que.push(node(xx,yy,cur+1));} } }int main(){ int t;cin>>t;while(t--){int i,j;ok=0;memset(vis,-1,sizeof vis);memset(visited,0,sizeof visited);cin>>r>>c; for (i=1;i<=r;i++)scanf("%s",tm[i]+1);while(!que.empty()) que.pop();for (i=1;i<=r;i++){for (j=1;j<=c;j++){if (tm[i][j]=='J')  {stx=i,sty=j;}if (tm[i][j]=='F') { que.push(node(i,j,0)); vis[i][j]=0;   } }}fire();  node st(stx,sty,0);while(!q.empty()) q.pop();q.push(st);int flag=-1;visited[stx][sty]=1;while(!q.empty()){node tp=q.front();q.pop();for (int i=0;i<4;i++){int xx=tp.x+dx[i];int yy=tp.y+dy[i];if (!(xx>=1&&xx<=r&&yy>=1&&yy<=c)){flag=tp.t;break;}if (tm[xx][yy]=='#')continue;if (vis[xx][yy]!=-1&&vis[xx][yy]<=tp.t+1)continue;if (visited[xx][yy])continue;q.push(node(xx,yy,tp.t+1));visited[xx][yy]=1;} if (flag!=-1)break;}if (flag!=-1) printf("%d\n",flag+1);else printf("IMPOSSIBLE\n");}return 0;}

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