Borg Maze(MST & bfs)

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<queue>
 4 using namespace std;
 5 int col , row , cnt;
 6 char maze[200][200] ;
 7 int l[200][200] ;
 8 int vis[200][200] ;
 9 int map[200][200] ;
10 const int inf = 0x3f3f3f3f ;
11 int move[][2] = {1 , 0 , 0 , 1 , -1 , 0 , 0 , -1} ;
12 
13 void bfs (int sx , int sy)
14 {
15     queue <pair <int , int> > q ;
16     while (!q.empty ())
17         q.pop () ;
18     memset (vis , -1 , sizeof(vis)) ;
19     vis[sx][sy] = 0 ;
20     q.push (make_pair(sx , sy)) ;
21     while (!q.empty ()) {
22         pair <int , int> k = q.front () ;
23         q.pop () ;
24         if (l[k.first][k.second] != -1)
25             map [l[sx][sy]] [l[k.first][k.second]] = vis [k.first][k.second] ;
26         for (int i = 0 ; i < 4 ; i++) {
27             int tx = k.first + move[i][0] ;
28             int ty = k.second + move[i][1] ;
29             if (maze[tx][ty] == ‘#‘ || vis[tx][ty] != -1)
30                 continue ;
31             vis[tx][ty] = vis[k.first][k.second] + 1 ;
32             q.push (make_pair(tx , ty)) ;
33         }
34     }
35 }
36 
37 void prim ()
38 {
39     int p[200] , d[200] ;
40     for (int i = 1 ; i < cnt ; i++) {
41         d[i] = map[1][i] ;
42         p[i] = 1 ;
43     }
44     d[1] = 0 ;
45     int ans = 0 ;
46     for (int i = 1 ; i < cnt - 1 ; i++) {
47         int minc = inf , k ;
48         for (int j = 1 ; j < cnt ; j++) {
49             if (d[j] && d[j] < minc) {
50                 minc = d[j] ;
51                 k = j ;
52             }
53         }
54         d[k] = 0 ;
55         for (int j = 1 ; j < cnt ; j++) {
56             if (d[j] && d[j] > map[k][j]) {
57                 d[j] = map[k][j] ;
58                 p[j] = k ;
59             }
60         }
61         ans += minc ;
62     }
63     printf ("%d\n" , ans) ;
64 }
65 
66 int main ()
67 {
68    // freopen ("a.txt" , "r" , stdin) ;
69     int T ;
70     scanf ("%d" , &T) ;
71     while (T--) {
72         scanf ("%d%d" , &col , &row) ;
73         gets(maze[0]) ;
74         int tol = 1 ;
75         memset (l , -1 , sizeof(l)) ;
76         for (int i = 0 ; i < row ; i++) {
77             gets (maze[i]) ;
78             for (int j = 0 ; j < col ; j++) {
79                 if (maze[i][j] == ‘A‘ || maze[i][j] == ‘S‘) {
80                     l[i][j] = tol++ ;
81                 }
82             }
83         }
84         for (int i = 0 ; i < row ; i++) {
85             for (int j = 0 ; j < col ; j++) {
86                 if (l[i][j] != -1) {
87                     bfs (i , j);
88                 }
89             }
90         }
91         cnt = tol ;
92         prim () ;
93     }
94     return 0 ;
95 }