hdu--2576

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

16 6 2...Y.....#...#.......#.....#....#G#.                

 

Sample Output

7 

 

解题思路：和迷宫题类似，题目大意为从Y出发是否能走到G的位置，如果可以输出最少时间，否则输出Please give me another chance!，但注意的时，当走到石头的地方时间为k的倍数时，是可以通过的，否则不能通过。对于Bfs题目最难的就是什么时候该标记的问题，这里只有当走过的地方为K的倍数时，才标记。

提供一组测试数据：

2 4 5

..Y#

.##G

6

代码如下：

#include<stdio.h>#include<queue>#include<string.h>#define INF 0x3f3f3f3fusing namespace std;int n,m,k;char mark[110][110];int vist[110][110];int dis[4][2]={{-1,0},{1,0},{0,1},{0,-1}};int ans;struct stu{int x,y,t;};stu s,e;void bfs(){memset(vist,0,sizeof(vist));//vist[s.x][s.y]=1;stu temp;stu vt;queue<stu>q;while(!q.empty()){q.pop();}q.push(s);while(!q.empty()){vt=q.front();q.pop();for(int i=0;i<4;i++){temp.x=vt.x+dis[i][0];temp.y=vt.y+dis[i][1];temp.t=vt.t+1;if(temp.x>=0&&temp.x<n&&temp.y>=0&&temp.y<m){if(!vist[temp.x][temp.y])     {if(mark[temp.x][temp.y]!='#'||temp.t%k==0){  if(temp.x==e.x&&temp.y==e.y)  {ans=temp.t;return;  }  else { if(temp.t%k==0)vist[temp.x][temp.y]=1;q.push(temp); }    }      }    }}}}int main(){int t;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&k);     for(int i=0;i<n;i++)     {     getchar();     for(int j=0;j<m;j++)     {     scanf("%c",&mark[i][j]);         if(mark[i][j]=='Y')         {         s.x=i;         s.y=j;         s.t=0;         }         else if(mark[i][j]=='G')         {         e.x=i;         e.y=j;         } }     }     if(s.x==e.x&&s.y==e.y)     {     printf("0\n");     continue;     }     ans=INF;         bfs();         if(ans>=INF)    printf("Please give me another chance!\n");    else    printf("%d\n",ans);}return 0;}