HDU - 2612 Find a way（宽搜）

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探讨了在一张网格地图上，两个角色如何找到能够使他们到达同一间肯德基餐厅总时间最短的路径。使用BFS算法分别计算两角色到达所有肯德基位置的时间，并找出最优解。

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
Input
The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’ express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF
Output
For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.
Sample Input
4 4
Y.#@
…
.#…
@…M
4 4
Y.#@
…
.#…
@#.M
5 5
Y…@.
.#…
.#…
@…M.
#…#
Sample Output
66
88
66

题意：Y和M要一起去肯德基，多组输入，给你一个m*n的地图，’.‘表示道路，’@'表示肯德基（有多个），‘Y’和‘M’表示两人的起始位置，’#'不能走。现在问你两人到同一家肯德基所需时间之和最少为多少。（两人分别可以上下左右移动，每移动一次需要11分钟，且一定会有解）

题解：两边BFS分别求出两人到各个肯德基所需的时间，再计算最小值。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct cc{
	int x,y,step;
};
queue<cc>q1;
queue<cc>q2;
char map[205][205];
int vis1[205][205];
int vis2[205][205];
int dx[4]={1,0,0,-1};
int dy[4]={0,1,-1,0};
int m,n;
void bfs()
{
	while(!q1.empty())
	{
		cc u=q1.front(); q1.pop();
		for(int j=0;j<4;j++)
		{
			int mx=u.x+dx[j],my=u.y+dy[j];
			if(mx>=1&&mx<=n&&my>=1&&my<=m&&!vis1[mx][my])
			{
				if(map[mx][my]=='.'||map[mx][my]=='@')
				{
					q1.push((cc){mx,my,u.step+1});
					vis1[mx][my]=u.step+1;
				}
			}
		}
	}
	while(!q2.empty())
	{
		cc u=q2.front(); q2.pop();
		for(int j=0;j<4;j++)
		{
			int mx=u.x+dx[j],my=u.y+dy[j];
			if(mx>=1&&mx<=n&&my>=1&&my<=m&&!vis2[mx][my])
			{
				if(map[mx][my]=='.'||map[mx][my]=='@')
				{
					q2.push((cc){mx,my,u.step+1});
					vis2[mx][my]=u.step+1;
				}
			}
		}
	}
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(vis1,0,sizeof(vis1));
		memset(vis2,0,sizeof(vis2));
		while(!q1.empty()) q1.pop();
		while(!q2.empty()) q2.pop();
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				cin>>map[i][j];
				if(map[i][j]=='Y')
				{
					q1.push((cc){i,j,0});
					vis1[i][j]=1;
				}
				if(map[i][j]=='M')
				{
					q2.push((cc){i,j,0});
					vis2[i][j]=1;
				}
			}
		}
		bfs();
		int ans=1e9;
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				if(map[i][j]=='@'&&vis1[i][j]!=0&&vis2[i][j]!=0)
				{
					if(ans>vis1[i][j]+vis2[i][j])
					{
						ans=vis1[i][j]+vis2[i][j];
					}
				}
			}
		}
		printf("%d\n",ans*11);
	}
	return 0;
}