hdoj 2579 Dating with girls(2)【三重数组标记去重】

Dating with girls(2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2705    Accepted Submission(s): 759

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

 

题意：给一个迷宫Y是起点，G是终点，#是石头不能通过 .是路可以走，但是当走到的步数step%k==0时#全部变为路可以通过，问从起点到终点的最少步数

题解：此题不用标记走过的路，但要避免走的路径重复走，同一个路径只在同一个时刻走过（当再次走到这个点且step%k刚好再次与此时相同时 不可以走），用一个三维数组标记  

从起点到终点，bfs，不同的是对于图中的点，可能走多次，分别是在不同的时刻。
vis[ t%k ][ i ][ j ]=1:表示坐标（ i，j ）在 t %k 时刻走过，接下来再出现 t%k 就不用再走一次了。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define MAX 110
#define INF 0x7ffff
using namespace std;
int n,m,k;
char map[MAX][MAX];
int vis[MAX][MAX][MAX];
int b,e;
struct node
{
	int x,y;
	int step;
	friend bool operator <(node a,node b)
	{
		return a.step>b.step;
	}
};
int judge(int a,int b)
{
	if(a<0||a>=n||b<0||b>=m)
	    return 0;
	return 1;
}
void bfs()
{
	int move[4][2]={1,0,-1,0,0,1,0,-1};
	node beg,end;
	priority_queue<node>q;
	while(!q.empty())
	q.pop();
	memset(vis,0,sizeof(vis));
	beg.x=b;
	beg.y=e;
	beg.step=0;
	vis[0][b][e]=1;
	q.push(beg);
	while(!q.empty())
	{
		end=q.top();
		q.pop();
		if(map[end.x][end.y]=='G')
		{
			printf("%d
",end.step);
			return ;
		}
		for(int i=0;i<4;i++)
		{
			beg.x=end.x+move[i][0];
			beg.y=end.y+move[i][1];
			if(judge(beg.x,beg.y))
			{
				if(map[beg.x][beg.y]!='#')
				{
					beg.step=end.step+1;
					if(!vis[beg.step%k][beg.x][beg.y])
					{
						vis[beg.step%k][beg.x][beg.y]=1;
						q.push(beg);
					} 
				}
				else
				{
					beg.step=end.step+1;
					if(!vis[beg.step%k][beg.x][beg.y]&&beg.step%k==0)
					{
						vis[beg.step%k][beg.x][beg.y]=1;
						q.push(beg);
					}
				}
			}
		}
	}
	printf("Please give me another chance!
");
}
int main()
{
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&n,&m,&k);
		for(i=0;i<n;i++)
		scanf("%s",map[i]);
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='Y')
				{
					b=i;
					e=j;
				}
			}
		}
		bfs();
	}
	return 0;
}