HDU 2579/BFS/ Dating with girls(2)

题目链接

/*
题意是是传统的迷宫加上一个条件，墙壁在k的整倍数时刻会消失，那么求到达出口的最短时间。
关键点在于某个点最多被走k次，标记vis[x][y][time%k]即可。
*/
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100+5;
int vis[maxn][maxn][15];
int dir[4][2]={0,1,1,0,0,-1,-1,0};
char maze[maxn][maxn];
int sx,sy,ex,ey;
int n,m,k;
void init()
{
    memset(vis,0,sizeof(vis));
}
struct node
{
    int x;
    int y;
    int time;
};
int BFS()
{
    node now;
    now.x=sx;
    now.y=sy;
    now.time=0;
    queue<node>que;
    que.push(now);
    while(!que.empty())
    {
        now=que.front();
        que.pop();
        if(now.x==ex&&now.y==ey)
            return now.time;
        for(int i=0;i<4;i++)
        {
            int x=now.x+dir[i][0];
            int y=now.y+dir[i][1];
            int t=now.time+1;
            int mod=t%k;
            if(x>=1&&x<=n&&y>=1&&y<=m)
            {
                if(maze[x][y]!='#'&&vis[x][y][mod]==0)
                {
                    vis[x][y][mod]=t;
                    node next=(node){x,y,t};
                    que.push(next);
                }
                else if(maze[x][y]=='#'&&mod==0&&vis[x][y][mod]==0)
                {
                    vis[x][y][mod]=t;
                    node next=(node){x,y,t};
                    que.push(next);
                }
            }
        }
    }
    return -1;
}
int main ()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%s",maze[i]+1);
            for(int j=1;j<=m;j++)
                if(maze[i][j]=='Y')
                    sx=i,sy=j;
                else if(maze[i][j]=='G')
                    ex=i,ey=j;
        }
        int ans=BFS();
        if(ans==-1)
            printf("Please give me another chance!\n");
        else
            printf("%d\n",ans);

    }
    return 0;
}