hdu 2579 Dating with girls(2)【BFS+全文翻译】

Dating with girls(2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2821    Accepted Submission(s): 791


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

个人版本的全文翻译（提供给英语不好的小伙伴和一些非英语考生0.0）：

如果你解决了Dating with girls(1).辣么我觉得你也可以解决这个问题。这个问题也是关于约妹纸。现在你在一个迷宫里并且有一个妹纸你想要约，如果你可以找到妹纸，那么你就可以约她，否则她就会约其他的男孩，多么蛋疼。。。

这个迷宫是非常奇怪的，迷宫里有很多石头。这些石头会在t时刻消失，t的定义是k的倍数，在其他时候石头是依然存在的.

这里保证只有. # Y G几种字符在迷宫的地图中，.只带你可以直接走的到地方，#表示石头，Y表示你的初始位子，G表示妹纸，保证只有一个Y和一个G，每一秒你可以向左走，向右走，向上走，或者向下走（翻译到这里大家应该懂了，不会有原地不动的情况哦~~~~~~）

思路：

因为有很多点可以重复走，所以我们要开三维数组vis【】【】【】来判重，如果一维用来表示x，一维用来表示y，那么另外一维用来表示什么呢？当然是走的步数了，但是我们不知道一共最多能走多少步啊~所以我们这里对k进行求余，只要是不为0，都表示石头存在。剩下的部分直接BFS不要怂就好。

注意的点：

可以直接走的点有： G Y 和 “。”这里我第一发交的时候忘记了Y这个地方也可以重复走，所以WA了一发。

间接能走的点有：# 即当前步数%k==0的时候可以走。

最后提供两组可能用的上的数据：

2
2 3 9
##.
G#Y
2 4 4
Y.#G
#.##

AC代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct zuobiao
{
int x,y,output;
}now,nex;
int fx[4]={0,0,1,-1};
int fy[4]={1,-1,0,0};
char a[105][105];
int vis[105][105][12];
int n,m,k;
void bfs(int x,int y)
{
memset(vis,0,sizeof(vis));
now.x=x;
now.y=y;
now.output=0;
vis[x][y][0]=1;
queue<zuobiao >s;
s.push(now);
while(!s.empty())
{
now=s.front();
if(a[now.x][now.y]=='G')
{
printf("%d\n",now.output);
return ;
}
s.pop();
for(int i=0;i<4;i++)
{
nex.x=now.x+fx[i];
nex.y=now.y+fy[i];
nex.output=now.output+1;
//printf("%d %d %d %d %d\n",nex.x,nex.y,nex.output,nex.output%k,vis[nex.x][nex.y][nex.output%k]);
if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&vis[nex.x][nex.y][nex.output%k]==0)
{
if(a[nex.x][nex.y]=='.'||a[nex.x][nex.y]=='G'||a[nex.x][nex.y]=='Y')
{
//printf("yes\n");
vis[nex.x][nex.y][nex.output%k]=1;
s.push(nex);
}
if(a[nex.x][nex.y]=='#'&&nex.output%k==0)
{
//printf("yes\n");
vis[nex.x][nex.y][nex.output%k]=1;
s.push(nex);
}
}
}
}
printf("Please give me another chance!\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int sx,sy;
scanf("%d%d%d",&n,&m,&k);
for(int i=0;i<n;i++)
{
scanf("%s",a[i]);
for(int j=0;j<m;j++)
{
if(a[i][j]=='Y')
{
sx=i;
sy=j;
}
}
}
bfs(sx,sy);
}
}