杭电-2579Dating with girls(2)（bfs+三维标记）

Dating with girls(2)

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

Total Submission(s): 2676    Accepted Submission(s): 749

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

 

这个题就是一个普通的bfs，但是要注意一点，在一定的时刻，障碍物是会消失的，这样一来，我们有些点可以走多次，每次取余k值不同是就是不同的状态，这时候，是可以重复走的，所以要用到三维数组记录！

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int m,n,k,bx,by,ex,ey;
char  map[110][110];
int vis[110][110][11],mov[4][2]={0,1,0,-1,-1,0,1,0};
struct node{
	int x,y,step;
}ta,tb;
queue<node>q;
bool can(node x,node y)
{
	if((x.step+1)%k==0&&y.x>=0&&y.x<m&&y.y>=0&&y.y<n&&!vis[y.x][y.y][(x.step+1)%k])
	return true;
	
	if(y.x>=0&&y.y>=0&&y.x<m&&y.y<n&&!vis[y.x][y.y][(x.step+1)%k]&&map[y.x][y.y]!='#')
	return true;
	
	return false;
}
int bfs(int x,int y)
{
	int i;
	while(!q.empty())
	q.pop();
	ta.x=x;
	ta.y=y;
	ta.step=0;
	q.push(ta);
	while(!q.empty())
	{
		ta=q.front();
		q.pop();
		if(ta.x==ex&&ta.y==ey)
		return ta.step;
		for(i=0;i<4;i++)
		{
			tb.x=ta.x+mov[i][0];
			tb.y=ta.y+mov[i][1];
			if(can(ta,tb))
			{
				vis[tb.x][tb.y][(ta.step+1)%k]=1;
				tb.step=ta.step+1;
				q.push(tb);
			}
		}
	}
	return -1;
}
int main()
{
	int N,i,j;
	scanf("%d",&N);
	while(N--)
	{
		scanf("%d%d%d",&m,&n,&k);
		for(i=0;i<m;i++)
		{
			scanf("%s",map[i]);
			for(j=0;j<n;j++)
			{
				if(map[i][j]=='Y')
				{
					bx=i;
					by=j;
				}
				if(map[i][j]=='G')
				{
					ex=i;
					ey=j;
				}
			}
		}
		memset(vis,0,sizeof(vis));
		vis[bx][by][0]=1;
		int t=bfs(bx,by);
		if(t==-1)
		printf("Please give me another chance!\n");
		else
		printf("%d\n",t);
	}
	return 0;
}