HDU 杭电2579 Dating with girls(2) 【BFS】

http://acm.hdu.edu.cn/showproblem.php?pid=2579

Problem Description

If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find the girl, then you can date with the girl.Else the girl will date with other boys. What a pity! 
The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there. 
There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move left, right, up or down.

 

Input

The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).
The next r line is the map’s description.

 

Output

For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".

 

Sample Input

   
   
   
   

    
    
    
    1
6 6 2
...Y..
...#..
.#....
...#..
...#..
..#G#.
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7
   
   
   
   

 

//标准的BFS，由于走过的路可以重复k次，为的是担心出现这种情况

//一开始我在比赛的时候没有注意到这种情况，所以又是一直WA

/*

Input：

1

6 6 3

......

...#..

.#....

...#..

...#Y.

..#G##

Output：

4

*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define MAXN 110
using namespace std;
struct node
{
    int x,y;
    int time;
    /*friend bool operator < (node a,node b)
    {
        return a.time>b.time;
    }*/
};
int row,column,k;
char map[MAXN][MAXN];
node s,e;
int x[4]={1,-1,0,0};
int y[4]={0,0,1,-1};
bool vis[MAXN][MAXN][20];
bool judge(int x,int y,int time)
{
    if(x>row||x<=0||y>column||y<=0||vis[x][y][time%k]) return 0;
    if(map[x][y]=='#')
    {
        if(time%k==0) return 1;
        return 0;
    }
    return 1;
}
int BFS()
{
    node pos,next;
    memset(vis,0,sizeof(vis));
    queue<node>q;
    vis[s.x][s.y][0] = 1;
    s.time = 0;
    q.push(s);    
    while(!q.empty())
    {
        pos=q.front();
        q.pop();
        for(int i=0;i<4;++i)
        {
            next.x=pos.x+x[i];
            next.y=pos.y+y[i];
            next.time = pos.time+1;
            if(judge(next.x,next.y,next.time))
            {
                if(next.x==e.x&&next.y==e.y) return next.time;
                vis[next.x][next.y][next.time%k]=1;
                q.push(next);
            }
        }
    }
    return -1;
}
int main()
{
    int T;
    int i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&row,&column,&k);
        getchar();
        for(i=1;i<=row;++i)
        {
            for(j=1;j<=column;++j)
            {
                scanf("%c",map[i]+j);
                if(map[i][j]=='G')
                {
                    e.x=i;e.y=j;
                }
                else if(map[i][j]=='Y')
                {
                    s.x=i,s.y=j;
                }
            }
            getchar();
        }
        int res=BFS();
        if(res==-1) printf("Please give me another chance!\n");
        else printf("%d\n",res);
    }
    return 0;
}