CCF_201604-4_游戏

bfs，首先记录危险方格的时间，因为100时间之后，所有方格均不危险，所以最短时间的最大值为300，记录每个坐标每个时间的状态，直接bfs即可。

#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;

struct point{
    int x,y,time;
}start;

int vis[][][] = {},a[][][] = {},dir[][] = {{-,},{,},{,-},{,}};

int main()
{
    int n,m,t;
    scanf("%d%d%d",&n,&m,&t);
    while(t--)
    {
        int x,y,start,stop;
        scanf("%d%d%d%d",&x,&y,&start,&stop);
        a[x][y][] = start;
        a[x][y][] = stop;
    }
    start.x = ;
    start.y = ;
    start.time = ;
    queue<point> q;
    q.push(start);
    while(!q.empty())
    {
        int x = q.front().x,y = q.front().y,time = q.front().time;
        if(x == n && y == m)
        {
            printf("%d\n",time);
            return ;
        }
        q.pop();
        for(int i = ;i < ;i++)
        {
            int xx = x+dir[i][],yy = y+dir[i][],timee = time+;
            if(xx <  || xx > n || yy <  || yy > m || timee>=a[xx][yy][]&&timee<=a[xx][yy][] || vis[xx][yy][timee])    continue;
            point temp;
            temp.x = xx;
            temp.y = yy;
            temp.time = timee;
            q.push(temp);
            vis[xx][yy][timee] = ;
        }
    }
}