运动员最佳匹配问题

Posted 2020-09-10 蒟蒻zht的博客

洛谷传送门

带权二分图最大权完美匹配。

裸的km算法。

注意开long long。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 
 5 using namespace std;
 6 
 7 const long long INF = 99999999999999999;
 8 int n, match[21];
 9 long long p[21][21], q[21][21], love[21][21], ex_boy[21], ex_girl[21], slack[21];
10 bool vis_boy[21], vis_girl[21];
11 
12 bool find(int i)
13 {
14     int j;
15     long long gap;
16     vis_girl[i] = 1;
17     for(j = 1; j <= n; j++)
18     {
19         if(vis_boy[j]) continue;
20         gap = ex_girl[i] + ex_boy[j] - love[i][j];
21         if(gap == 0)
22         {
23             vis_boy[j] = 1;
24             if(match[j] == -1 || find(match[j]))
25             {
26                 match[j] = i;
27                 return 1;
28             }
29         }
30         else slack[j] = min(slack[j], gap);
31     }
32     return 0;
33 }
34 
35 long long KM()
36 {
37     int i, j;
38     long long d, ret = 0;
39     memset(match, -1, sizeof(match));
40     for(i = 1; i <= n; i++)
41      for(j = 1; j <= n; j++)
42       ex_girl[i] = max(ex_girl[i], love[i][j]);
43     for(i = 1; i <= n; i++)
44     {
45         for(j = 1; j <= n; j++) slack[j] = INF;
46         while(1)
47         {
48             memset(vis_boy, 0, sizeof(vis_boy));
49             memset(vis_girl, 0, sizeof(vis_girl));
50             if(find(i)) break;
51             d = INF;
52             for(j = 1; j <= n; j++)
53              if(!vis_boy[j])
54               d = min(d, slack[j]);
55             for(j = 1; j <= n; j++)
56             {
57                 if(vis_boy[j]) ex_boy[j] += d;
58                 if(vis_girl[j]) ex_girl[j] -= d;
59             }
60         }
61     }
62     for(i = 1; i <= n; i++) ret += love[match[i]][i];
63     return ret;
64 }
65 
66 int main()
67 {
68     int i, j;
69     scanf("%d", &n);
70     for(i = 1; i <= n; i++)
71      for(j = 1; j <= n; j++)
72       scanf("%lld", &p[i][j]);
73     for(i = 1; i <= n; i++)
74      for(j = 1; j <= n; j++)
75       scanf("%lld", &q[i][j]);
76     for(i = 1; i <= n; i++)
77      for(j = 1; j <= n; j++)
78       love[i][j] = p[i][j] * q[j][i];
79     printf("%lld", KM());
80     return 0;
81 }

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