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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[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;        }    }}

 

CCF_201604-4_游戏