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

Output

Sample Input

Sample Output

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cstdlib>#include<cmath>#include<algorithm>#include<queue>#include<vector>#include<set>using namespace std;int n,m,k;int x_s,y_s;int x_e,y_e;char map[111][111];int step[10][111][111];int dir[4][2]={0,1, 1,0, 0,-1, -1,0};struct node{	int f,x,y;};int BFS(){	queue<node>q;	node now,next;	int i;	memset(step,-1,sizeof(step));	now.f=0;	now.x=x_s;	now.y=y_s;	q.push(now);	step[now.f][now.x][now.y]=0;	while(!q.empty())	{		now=q.front();		q.pop();		if(now.x==x_e && now.y==y_e)	return step[now.f][now.x][now.y];		for(i=0;i<4;i++)		{			next.x=now.x+dir[i][0];			next.y=now.y+dir[i][1];			next.f=(now.f+1)%k;			if(next.x<0 || next.x>=n || next.y<0 || next.y>=m)	continue;			if(map[next.x][next.y]==‘#‘ && next.f)				continue;			if(step[next.f][next.x][next.y]!=-1)				continue;			step[next.f][next.x][next.y]=step[now.f][now.x][now.y]+1;			q.push(next);		}	}	return -1;}int main(){	int T;	int i,l;	int ans;	scanf("%d",&T);	while(T--)	{		scanf("%d%d%d",&n,&m,&k);		for(i=0;i<n;i++)		{			scanf("%s",map[i]);			for(l=0;map[i][l];l++)			{				if(map[i][l]==‘Y‘)	{x_s=i;y_s=l;map[i][l]=‘.‘;}				if(map[i][l]==‘G‘)	{x_e=i;y_e=l;map[i][l]=‘.‘;}			}		}		ans=BFS();		if(ans==-1)	printf("Please give me another chance!\n");		else		printf("%d\n",ans);	}	return 0;}