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  • UVA 11624 Fire!

    数据:

    3
    4 4
    ####
    #JF#
    #..#
    #..#
    3 3
    FFF
    FJF
    FFF
    3 3
    ...
    .J.
    ...

    嗯,被第三个数据卡啦半天....

    两次BFS  预处理一下每个点最早着火的时间.

    /* ***********************************************
    Author        :pk29
    Created Time  :2015/8/19 19:26:13
    File Name     :4.cpp
    ************************************************ */
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <iomanip>
    #include <list>
    #include <deque>
    #include <stack>
    #define ull unsigned long long
    #define ll long long
    #define mod 90001
    #define INF 0x3f3f3f3f
    #define maxn 10000+10
    #define cle(a) memset(a,0,sizeof(a))
    const ull inf = 1LL << 61;
    const double eps=1e-5;
    using namespace std;
    struct node{
        int x,y,dist;
    };
    bool cmp(int a,int b){
        return a>b;
    }
    int dir[4][2]={1,0,0,1,0,-1,-1,0};
    char mp[1100][1100];
    int vis[1100][1100],val[1100][1100];
    int n,m,sx,sy,fx,fy;
    queue<node>p;
    void bfs1(){
        cle(vis);
        node u,v;
        while(!p.empty()){
            u=p.front(),p.pop();
            for(int i=0;i<4;i++){
                int nx=u.x+dir[i][0];
                int ny=u.y+dir[i][1];
                if(!vis[nx][ny]&&nx<=n&&ny>=1&&nx>=1&&ny<=m&&mp[nx][ny]=='.'){
                    v.x=nx,v.y=ny,v.dist=u.dist+1;
                    vis[nx][ny]=1;
                    val[nx][ny]=v.dist;
                    p.push(v);
                }
            }
        }
    }
    queue<node>q;
    void bfs2(){
        int mark=0;
        cle(vis);
        while(!q.empty())q.pop();
        node u,v;
        u.x=sx,u.y=sy,u.dist=0;
        q.push(u);
        vis[u.x][u.y]=1;
        while(!q.empty()){
            u=q.front(),q.pop();
            if(u.x==n||u.x==1||u.y==1||u.y==m){
                mark=1;printf("%d
    ",u.dist+1);break;
            }
            for(int i=0;i<4;i++){
                int nx=u.x+dir[i][0];
                int ny=u.y+dir[i][1];
                if(vis[nx][ny]==0&&nx<=n&&ny>=1&&nx>=1&&ny<=m&&mp[nx][ny]=='.'){
                    v.dist=u.dist+1;
                    if(v.dist<val[nx][ny]){
                        v.x=nx,v.y=ny;
                        q.push(v);
                        vis[nx][ny]=1;
                    }
                }
            }
        }
        if(!mark)printf("IMPOSSIBLE
    ");
    }
    int main()
    {
        #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        int t;
        cin>>t;
        while(t--){
            cin>>n>>m;
            cle(mp);
            memset(val,INF,sizeof val);
            for(int i=1;i<=n;i++)
                for(int j=1;j<=m;j++){
                    cin>>mp[i][j];
                    if(mp[i][j]=='J'){
                        sx=i,sy=j;
                    }
                    if(mp[i][j]=='F'){
                        node o;o.x=i,o.y=j,o.dist=0;
                         p.push(o);
                    }
                }
            bfs1();
            bfs2();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pk28/p/4743931.html
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