【题目描述】
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

【题目分析】
进过分析就是一个很简单的BFS，只不过需要对两个人分别进行BFS然后再依次对每个KFC判断，取步数和最少的。
需要注意的是如果初始化的时候步数为0，那么某个KFC的步数为0说明有一个人无法到达这个地方，需要判断，否则会错
【AC代码】

#include<cstdio>
#include<cstring>
#include<queue>
#include<climits>
using namespace std;const int drc[4][2]={{1,0},{-1,0},{0,1,},{0,-1}};
const int MAXN=205;
char map[MAXN][MAXN];
bool vis[MAXN][MAXN];
int a1[MAXN][MAXN];
int a2[MAXN][MAXN];
int n,m,ans;
int x1,x2,y1,y2;struct node
{int x,y;int step;
}t,p;
int cnt=0;void BFS1()
{int u,v;queue<node> q;p.x=x1; p.y=y1; p.step=0;memset(vis,0,sizeof(vis));memset(a1,0,sizeof(a1));vis[x1][y1]=1;q.push(p);while(!q.empty()){p=q.front(); q.pop();if(map[p.x][p.y]=='@'){a1[p.x][p.y]=p.step;}for(int i=0;i<4;i++){u=p.x+drc[i][0]; v=p.y+drc[i][1];if(u<0 || u>=n || v<0 || v>=m) continue;if(map[u][v]=='#' || vis[u][v]) continue;vis[u][v]=1;t.x=u; t.y=v; t.step=p.step+1;q.push(t);}}
}void BFS2()
{int u,v;queue<node> q;p.x=x2; p.y=y2; p.step=0;memset(vis,0,sizeof(vis));memset(a2,0,sizeof(a1));vis[x2][y2]=1;q.push(p);while(!q.empty()){p=q.front(); q.pop();if(map[p.x][p.y]=='@'){a2[p.x][p.y]=p.step;}for(int i=0;i<4;i++){u=p.x+drc[i][0]; v=p.y+drc[i][1];if(u<0 || u>=n || v<0 || v>=m) continue;if(map[u][v]=='#' || vis[u][v]) continue;vis[u][v]=1;t.x=u; t.y=v; t.step=p.step+1;q.push(t);}}
}int main()
{while(~scanf("%d%d",&n,&m)){memset(map,0,sizeof(map));for(int i=0;i<n;i++){scanf("%s",map[i]);for(int j=0;j<m;j++){if(map[i][j]=='Y'){x1=i; y1=j; map[i][j]='.';continue;}if(map[i][j]=='M'){x2=i; y2=j; map[i][j]='.';continue;}}}BFS1();BFS2();ans=INT_MAX;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(map[i][j]=='@' && a1[i][j]!=0 && a2[i][j]!=0)	//很重要{ans=min(ans,a1[i][j]+a2[i][j]);}}}printf("%d\n",ans*11);}return 0;
}