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

開始以為用雙向bfs，兩個人都走到的就直接輸出答案，但實際對於到當前點，雖然Y和M都是最短，但總路程不一定最短，所以可以持續更新，也可以最後遍歷一遍。

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 205;
int n,m;
char a[N][N];

int dx[4] = {-1,0,0,1};
int dy[4] = {0,-1,1,0}; 
int dis[3][N][N];
int ans;

struct node {
	int x,y,id;
}s1,s2;

void bfs() {
	queue <node> q;
	q.push(s1); q.push(s2);
	while(!q.empty()) {
		node cur = q.front(); q.pop();
		int x = cur.x,y = cur.y;
		if(dis[1][x][y] && dis[2][x][y] && a[x][y] == '@') {
			//printf("%d %d\n",x,y);//debuging
			//cout << (dis[1][x][y]+dis[2][x][y])*11 << endl;//debug dis1+dis2 not max(dis1,dis2);
			//break;
			ans = min(ans,dis[1][x][y]+dis[2][x][y]);
		}
		for(int i = 0;i < 4; i++) {
			int nx = x+dx[i],ny = y+dy[i],id = cur.id;
			if(nx < 1 || nx > n || ny < 1 || ny > m || a[nx][ny] == '#' || dis[id][nx][ny]) continue;//debug ny > m not < m 
			node pt; pt.x = nx,pt.y = ny,pt.id = id;
			q.push(pt); dis[id][nx][ny] = dis[id][x][y]+1;//debug +1 do not write
		}
	}
} 

int main() {
	while(cin >> n >> m) {
		ans = 1<<30;//debug init ans;
		for(int i = 1;i <= n; i++) {
			for(int j = 1;j <= m; j++) {
				cin >> a[i][j]; dis[1][i][j] = dis[2][i][j] = 0;//debug not init dis
				if(a[i][j] == 'Y') s1.x = i,s1.y = j,s1.id = 1;
				if(a[i][j] == 'M') s2.x = i,s2.y = j,s2.id = 2; 
			}
		}
		bfs();
		cout << ans*11 << endl;
	}
	return 0;
}