hoj 2543 (费用流 拆边）

http://acm.hit.edu.cn/hoj/problem/view?id=2543

1.将原图中的每条边(u, v)拆成两条:(u, v, Ci, 0), (u, v, ∞, Ei)

2.购买的每个石头的费用P加一条 (S, 1, inf, P)的边。

3.总的能够花费的费用C可以在我们求最小费用路的时候判断。

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

 using namespace std;
 typedef long long LL;
 const int maxn = 1e3 + ;
 const int maxm = 1e5+ ;
 const int inf = 0x3f3f3f3f;

 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;
     }
     void mcmf(int S, int T, int ct)
     {
         sumFlow = ;
         LL minFlow = , minCost = ;
         while (spfa(S, T))
         {
             minFlow = inf + ;
             for (int i = pre[T]; i + ; i = pre[ p[i ^ ].v ])
                 minFlow = min(minFlow,(LL)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;
             if (minCost > ct)
             {
                 sumFlow -= ceil((minCost - ct) * 1.0 / dis[T]);
                 break;
             }
         }
     }
     void build(int nt, int mt, int ct, int pt)
     {
         init(nt);
         int u, v, c, w;
         addEdge(, , inf, pt);
         while (mt--)
         {
             scanf("%d%d%d%d", &u, &v, &c, &w);
             u++, v++;
             addEdge(u, v, c, ); addEdge(u, v, inf, w);
             swap(u, v);
             addEdge(u, v, c, ); addEdge(u, v, inf, w);
         }
     }
     void solve(int nt, int mt, int ct, int pt)
     {
         build(nt, mt, ct, pt);
         mcmf(, , ct);
         printf("%d\n", sumFlow);
     }
 }my;
 int main()
 {
     int tcase, n, m, c, p;
     scanf("%d", &tcase);
     while (tcase--)
     {
         scanf("%d%d%d%d",&n, &m, &c, &p);
         my.solve(n, m, c, p);
     }
     return ;
 }