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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.
[email protected] 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

思路：兩次BFS存下對每個kfc的最短距離，之後兩兩相加取min

代碼：

 1 #include "cstdio"
 2 #include "stdlib.h"
 3 #include "iostream"
 4 #include "algorithm"
 5 #include "string"
 6 #include "cstring
 
"
 7 #include "queue"
 8 #include "cmath"
 9 #include "vector"
10 #include "map"
11 #include "set"
12 #define mj
13 #define db double
14 #define ll long long
15 using namespace std;
16 const int N=1e8+2;
17 const int mod=1e9+7;
18 //const ll inf=1e16+10;
19 #define inf 0x3f3f3f
20 typedef pair<int,int> P;
21 int n,m;
22 char  s[300][300];
23 int d[300][300],k[205][205];
24 int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
25 int t[2][205*205];
26 int bfs(int sx,int sy,int id)
27 {
28     queue<P> q;
29     for(int i=0;i<205;i++){
30         for(int j=0;j<205;j++){
31             d[i][j]=N;
32         }
33     }
34     q.push(P(sx,sy));
35     d[sx][sy]=0;
36     while(q.size()){
37         P p;
38         p=q.front(),q.pop();
39         for(int i=0;i<4;i++){
40             int nx=p.first+dx[i],ny=p.second+dy[i];
41             if(0<=nx&&nx<n&&0<=ny&&ny<m&&s[nx][ny]!=‘#‘&&d[nx][ny]==N){
42                 d[nx][ny]=d[p.first][p.second]+1;
43                 if(s[nx][ny]==‘@‘) t[id][k[nx][ny]]=d[nx][ny];//到該點的距離
44                 q.push(P(nx,ny));
45             }
46         }
47     }
48     return 0;
49 }
50 int main()
51 {
52     int xx[3],yy[3];
53     while(scanf("%d%d",&n,&m)==2){
54         memset(t,inf, sizeof(t));
55         int c=0,cnt=0,ma=N;
56         for(int i=0;i<n;i++){
57             scanf("%s",s[i]);
58             for(int j=0;j<m;j++){
59                 if(s[i][j]==‘Y‘) xx[c]=i,yy[c++]=j;
60                 else if(s[i][j]==‘M‘) xx[c]=i,yy[c++]=j;
61                 else if(s[i][j]==‘@‘){
62                     k[i][j]=cnt++;
63                 }
64             }
65         }
66         bfs(xx[0],yy[0],0),bfs(xx[1],yy[1],1);
67         for(int i=0;i<cnt;i++){
68             ma=min(t[0][i]+t[1][i],ma);
69 //            printf("%d %d\n",t[0][i],t[1][i]);
70         }
71         printf("%d\n",11*ma);
72     }
73     return 0;
74 }

hdu 2612 多終點BFS