最小费用最大流模板

 1 bool spfa()
 2 {
 3      fill(vs,vs+n+2,false);
 4      fill(d,d+n+2,inf);
 5      fill(father,father+n+2,-1);
 6      queue<int>q;
 7      d[0]=0;
 8      q.push(0);
 9      while (!q.empty())
10      {
11          int u=q.front();
12          q.pop();
13          vs[u]=false;
14          for (int i=head[u];i!=-1;i=eage[i].next)
15          {
16              int v=eage[i].v;
17              if (eage[i].cap&&d[v]>d[u]+eage[i].cost)
18              {
19                  d[v]=d[u]+eage[i].cost;
20                  father[v]=i;
21                  if (!vs[v])
22                  {
23                      vs[v]=true;
24                      q.push(v);
25                  }
26              }
27          }
28      }
29      if (father[n+1]==-1) return false;
30      return true;
31 }
32 
33 int solve()
34 {
35     int ans=0;
36     while (spfa())
37     {
38         ans+=d[n+1];
39         int u=n+1;
40         while (u!=0)
41         {
42             int i=father[u];
43             eage[i].cap--;
44             eage[i^1].cap++;
45             u=eage[i].u;
46         }
47     }
48     return ans;
49 }