FZU 2150 Fire Game 枚举BFS

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考虑的n，和m较小，可以枚举2个放火的位置，然后BFS，

#include<iostream>#include<cstdio>#include<math.h>#include<algorithm>#include<map>#include<set>#include<stack>#include<queue>#include<string.h>#include<cstring>#include<vector>using namespace std;#define INF 0x3f3f3f3f#define mem(x,y) memset(x,y,sizeof(x))typedef long long LL;#define MX 1111#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef pair<int,int> PII;struct Point{    int x,y,t;    Point(int x,int y,int t):x(x),y(y),t(t) {}};int n,m;int to[4][2]= {1,0,-1,0,0,1,0,-1};char _map[11][11];bool vis[11][11];int bfs(int x1,int y1,int x2,int y2){    mem(vis,0);    queue<Point> my;    my.push(Point(x1,y1,0));    my.push(Point(x2,y2,0));    vis[x1][y1]=vis[x2][y2]=1;    int ans=0;    while(!my.empty())    {        Point u=my.front();        my.pop();        for(int i=0; i<4; i++)        {            int xx=u.x+to[i][0],yy=u.y+to[i][1],tt=u.t+1;            if(xx<1||xx>n||yy<1||yy>m||vis[xx][yy]||_map[xx][yy]=='.') continue;            ans=max(ans,tt);            vis[xx][yy]=1;            my.push(Point(xx,yy,tt));        }    }    for(int i=1; i<=n; i++)        for(int j=1; j<=m; j++) if(_map[i][j]=='#'&&vis[i][j]==0) return INF;    return ans;}int main(){   // freopen("input.txt","r",stdin);    int T;////    cin>>T;    int tt=0;    while(T--)    {        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++) scanf("%s",_map[i]+1);

        //一开始也可以先bfs下判断联通块的个数，大于2的直接输出-1就好        int ans=INF;        for(int x1=1; x1<=n; x1++)            for(int y1=1; y1<=m; y1++)                for(int x2=x1; x2<=n; x2++)//这里加上一个强行减枝，x2要>=x1，不然会超时。                    for(int y2=1; y2<=m; y2++) if(_map[x1][y1]=='#'&&_map[x2][y2]=='#')ans=min(ans,bfs(x1,y1,x2,y2));        printf("Case %d: %d\n",++tt,ans==INF?-1:ans);    }    return 0;}

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