<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
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. 

</blockquote>
<h2 id="input">Input</h2>
<blockquote>
The input contains multiple test cases. 
Each test case include, first two integers n, m. (2&lt;=n,m&lt;=200). 
Next n lines, each line included m character. 
‘Y’ express yifenfei initial position. 
‘M’ express Merceki initial position. 
‘#’ forbid road; 
‘.’ Road. 
‘@’ KCF
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
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.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
4 4 
Y.#@ 
.... 
.#.. 
@..M 
4 4 
Y.#@ 
.... 
.#.. 
@#.M 
5 5 
Y..@. 
.#... 
.#... 
@..M. 
#...#
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
66 
88 
66
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
有两个起点的BFS，每个KFS是一个出口，分别计算出每个出口距离两个入口的距离，计算其最小值
要特别注意：KFC有可能不能到达
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;map&gt;
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o&lt;n;o++)

const int maxn = 205;
int n,m;
char Map[maxn][maxn];
const int delta[] = {1,-1,0,0};

struct point {
 int x,y;
 point() {
 x = y = -1;
 }
 point(int a,int b) {
 x = a;
 y = b;
 }
};
int num;


int BFS(point s,int (&amp;dis)[maxn][maxn]) {
 memset(dis,-1,sizeof(dis));
 queue&lt;point&gt; Q;

 Q.push(s);
 dis[s.x][s.y] = 0;

 while (!Q.empty()) {
 int x = Q.front().x;
 int y = Q.front().y;
 Q.pop();

 REP(4) {
 int xx = x + delta[o];
 int yy = y + delta[3 - o];

 if (xx &lt; 0 || xx &gt;= n || yy &lt; 0 || yy &gt;= m)
 continue;
 if (Map[xx][yy] == &#39;#&#39;)
 continue;
 if (dis[xx][yy] == -1) {
 dis[xx][yy] = dis[x][y] + 1;
 Q.push(point(xx,yy));
 }
 }
 }
 return -1;
}

bool Do() {
 if (scanf(&#34;%d%d&#34;,&amp;n,&amp;m) == EOF)
 return false;
 point s1,s2;
 num = 0;
 point v[maxn*maxn];
 for (int i = 0;i &lt; n;i++)
 for (int j = 0;j &lt; m;j++) {
 scanf(&#34;\n%c&#34;,&amp;Map[i][j]);
 if (Map[i][j] == &#39;Y&#39;)
 s1 = point(i,j);
 if (Map[i][j] == &#39;M&#39;)
 s2 = point(i,j);
 if (Map[i][j] == &#39;@&#39;) 
 v[num++] = point(i,j);
 }

 int dis1[maxn][maxn];
 int dis2[maxn][maxn];

 BFS(s1,dis1);
 BFS(s2,dis2);

 int Min = 100000;
 for (int i = 0;i &lt; num;i++)
 if(dis1[v[i].x][v[i].y]!=-1&amp;&amp; dis2[v[i].x][v[i].y]!=-1)
 Min = min(Min,dis1[v[i].x][v[i].y]+ dis2[v[i].x][v[i].y]);

 printf(&#34;%d\n&#34;,Min * 11);

 return true;
}

int main() {
 while (Do());
 return 0;
}
</pre>

HDU 2612.Find a way