hdu 3667 (拆边 mcmf)

注意题目中 边的容量 <= 5.可以把费用权值 a *f ^2化归成 a * f2, 即第一条边费用为 1 * a, 第二条 为 (4 - 1) * a, 第三条为 (9  - 4) * a。。。。。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>

 using namespace std;

 const int maxn = ;
 const int maxm = ;
 const int inf = 0x3f3f3f3f;
 const int ww[] = {, , , , };

 struct MCMF
 {
     struct Edge
     {
         int v, c, w, next;
     }p[maxm << ];
     int e, head[maxn], dis[maxn], pre[maxn], cnt[maxn], sumFlow, n;
     bool vis[maxn];
     void init(int nt)
     {
         e = ; n = nt;
         memset(head, -, sizeof(head[]) * (n + ));
     }
     void addEdge(int u, int v, int c, int w)
     {
         p[e].v = v; p[e].c = c; p[e].w = w; p[e].next = head[u]; head[u] = e++;
         swap(u, v);
         p[e].v = v; p[e].c = ; p[e].w =-w; p[e].next = head[u]; head[u] = e++;
     }
     bool spfa(int S, int T)
     {
         queue <int> q;
         for (int i = ; i <= n; ++i)
             vis[i] = cnt[i] = , pre[i] = -, dis[i] = inf;
         vis[S] = ; dis[S] = ;
         q.push(S);
         while (!q.empty())
         {
             int u = q.front(); q.pop();
             vis[u] = ;
             for (int i = head[u]; i + ; i = p[i].next)
             {
                 int v = p[i].v;
                 if (p[i].c && dis[v] > dis[u] + p[i].w)
                 {
                     dis[v] = dis[u] + p[i].w;
                     pre[v] = i;
                     if (!vis[v])
                     {
                         q.push(v);
                         vis[v] = ;
                         if (++cnt[v] > n) return ;
                     }
                 }
             }
         }
         return dis[T] != inf;
     }
     int mcmf(int S, int T)
     {
         sumFlow = ;
         int minFlow = , minCost = ;
         while (spfa(S, T))
         {
             minFlow = inf + ;
             for (int i = pre[T]; i + ; i = pre[ p[i ^ ].v ])
                 minFlow = min(minFlow, p[i].c);
             sumFlow += minFlow;
             for (int i = pre[T]; i + ; i = pre[ p[i ^ ].v ])
                 p[i].c -= minFlow, p[i ^ ].c += minFlow;
             minCost += dis[T] * minFlow;
         }
         return minCost;
     }
     void build(int nt, int mt, int kt)
     {
         init(nt);
         addEdge(, , kt, );
         int u, v, c, w;
         for (int i = ; i < mt; ++i)
         {
             scanf("%d%d%d%d", &u, &v, &w, &c);
             for (int j = ; j < c; ++j)
                 addEdge(u, v, , w * ww[j]);
         }
     }
     void solve(int nt, int mt, int kt)
     {
         build(nt, mt, kt);
         int ans = mcmf(, n);
     //    cout <<"sumFlow = "  << sumFlow << " cost = " << ans << endl;
         if (sumFlow != kt)
             printf("-1\n");
         else
             printf("%d\n", ans);
     }
 }my;

 int main()
 {
     int n, m, k;
     while (~scanf("%d%d%d", &n, &m, &k))
         my.solve(n, m, k);
     return ;
 }