【題目描述】
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

【題目分析】
進過分析就是一個很簡單的BFS，只不過需要對兩個人分別進行BFS然后再依次對每個KFC判斷，取步數和最少的。
需要注意的是如果初始化的時候步數為0，那么某個KFC的步數為0說明有一個人無法到達這個地方，需要判斷，否則會錯
【AC代碼】

#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;
}