CCF_201604-4_游戏

Posted 2020-08-06 冷暖知不知

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[105][105][305] = {0},a[105][105][2] = {0},dir[][2] = {{-1,0},{1,0},{0,-1},{0,1}};

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][0] = start;
        a[x][y][1] = stop;
    }
    start.x = 1;
    start.y = 1;
    start.time = 0;
    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 0;
        }
        q.pop();
        for(int i = 0;i < 4;i++)
        {
            int xx = x+dir[i][0],yy = y+dir[i][1],timee = time+1;
            if(xx < 1 || xx > n || yy < 1 || yy > m || timee>=a[xx][yy][0]&&timee<=a[xx][yy][1] || vis[xx][yy][timee])    continue;
            point temp;
            temp.x = xx;
            temp.y = yy;
            temp.time = timee;
            q.push(temp);
            vis[xx][yy][timee] = 1;
        }
    }
}