HDoj 2612 Find a way【bfs+キュー】

Find a way
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5401    Accepted Submission(s): 1823
Problem Description
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
2つの始点、複数の終点、2人は同じ終点に到着しなければならず、2つの配列を開いて、それぞれ1番目と2番目の人が終点に到着したステップ数を保存し、それから対応する終点で2人のステップ数を加算して、最小の

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define MAX 210
#define INF 0x3f3f3f
using namespace std;
int bu1[MAX][MAX];//         
int bu2[MAX][MAX];//         
int p;
char map[MAX][MAX];
int vis[MAX][MAX];
int n,m;
struct node
{
	int x,y;
	int step;
};
int MIN(int x,int y)
{
	return x<y?x:y;
}
void bfs(int x1,int y1,int p)
{
	memset(vis,0,sizeof(vis));
	int j,i,ok=0;
	int move[4][2]={0,1,0,-1,1,0,-1,0};
	queue<node>q;
	node beg,end;
	beg.x=x1;
	beg.y=y1;
	beg.step=0;
	q.push(beg);
	while(!q.empty())
	{
		end=q.front();
		q.pop();
		if(map[end.x][end.y]=='@')//  @        
		{
			if(p==1)
			bu1[end.x][end.y]=end.step;
			else
			bu2[end.x][end.y]=end.step;
		}
		for(i=0;i<4;i++)
		{
			beg.x=end.x+move[i][0];
			beg.y=end.y+move[i][1];
			if(!vis[beg.x][beg.y]&&0<=beg.x&&beg.x<n&&0<=beg.y&&beg.y<m&&map[beg.x][beg.y]!='#')
			{
				vis[beg.x][beg.y]=1;
				beg.step=end.step+11;
				q.push(beg);
			}
		}
	}
}
int main()
{
	int sum,j,i,t,k,x1,x2,y1,y2,min;
	int s[11000];
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		
		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')
				{
					x1=i;y1=j;
				}
				else if(map[i][j]=='M')
				{
					x2=i;y2=j;
				}
			}
		}
		memset(bu1,INF,sizeof(bu1));
	    bfs(x1,y1,1);
	    memset(bu2,INF,sizeof(bu2));
		bfs(x2,y2,2);
		min=INF;
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(bu1[i][j]!=INF&&bu2[i][j]!=INF)
				{
					min=MIN(bu1[i][j]+bu2[i][j],min);//           
				}
			}
		}
		printf("%d
",min);
	}
	return 0;      
}