1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<queue>
6 using namespace std;
7
8 char map[201][201];
9 int book1[205][205],book2[205][205];
10 int step1[205][205],step2[205][205];
11 int dir[4][2]={ {1,0},{0,1},{-1,0},{0,-1} };
12 int n,m,x1,y1,x2,y2;
13
14 struct note
15 {
16 int x;
17 int y;
18 int s;
19 }a,b;
20 queue<note>q;
21
22 void bfs1(int xx,int yy)
23 {
24 memset(book1,0,sizeof(book1));
25 while(!q.empty()) //清空
26 q.pop();
27 a.x= xx;
28 a.y= yy;
29 a.s= 0 ;
30 book1[a.x][a.y]=1;
31 step1[a.x][a.y]=0;
32 q.push(a);
33 while(!q.empty())
34 {
35 a=q.front();
36 q.pop();//及时出队
37 for(int i=0;i<4;i++)
38 {
39 b.x=a.x+dir[i][0];
40 b.y=a.y+dir[i][1];
41 if(map[b.x][b.y]=='#'||
42 book1[b.x][b.y]==1||
43 b.x<1||b.x>n||b.y<1||b.y>m)
44 continue;
45 book1[bx][by]= 1 ; // mark: walked
46 bs=a.s+ 11 ;
47 step1[bx][by]=bs; // record the number of steps to each position
48 q.push (b); // Enqueue
49 }
50 }
51 return ;
52 }
53
54 void bfs2( int xx, int yy)
55 {
56 memset(book2, 0 , sizeof (book2));
57 while (!q.empty())
58 q.pop();
59 a.x= xx;
60 a.y= yy;
61 a.s= 0 ;
62 book2[a.x][a.y]=1;
63 step2[a.x][a.y]=0;
64 q.push(a);
65 while(!q.empty())
66 {
67 a=q.front();
68 q.pop();
69 for(int i=0;i<4;i++)
70 {
71 b.x=a.x+dir[i][0];
72 b.y=a.y+dir[i][1];
73 if(map[b.x][b.y]=='#'||
74 book2[b.x][b.y]==1||
75 b.x<1||b.x>n||b.y<1||b.y>m)
76 continue;
77 book2[b.x][b.y]=1;
78 b.s=a.s+11;
79 step2[b.x][b.y]=b.s;
80 q.push(b);
81 }
82 }
83 return;
84 }
85
86 int main()
87 {
88 while(cin>>n>>m)
89 {
90 for(int i=1;i<=n;i++)
91 {
92 for(int j=1;j<=m;j++)
93 {
94 cin>>map[i][j];
95 if(map[i][j]=='Y')
96 {
97 x1=i; y1=j;
98 }
99 else if(map[i][j]=='M')
100 {
101 x2=i; y2=j;
102 }
103 }
104 }
105
106 bfs1(x1,y1);
107 bfs2(x2,y2);
108
109 int minn=0x3f3f3f3f;
110 for ( int i= 1 ;i<=n;i++ )
111 {
112 for ( int j= 1 ;j<=m;j++ )
113 {
114 if (map[i][j]== ' @ ' &&book1 [i][j]== 1 &&book2[i][j]== 1 ) //Be sure to check the book here (because someone may not be able to reach the place @
115 minn=min( minn, step1[i][j] + step2[i][j]);
116 }
117 }
118 cout<<minn<< endl;
119 }
120 }