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  • 最小费用最大流算法

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define INF 0x3f3f3f3f
     4 #define M(a, b) memset(a, b, sizeof(a))
     5 const int N = 1e3 + 5;
     6 struct Edge {
     7     int from, to, cap, flow, cost;
     8 };
     9 
    10 struct MCMF {
    11     int n, m;
    12     vector<Edge> edges;
    13     vector<int> G[N];
    14     int d[N], inq[N], p[N], a[N];
    15 
    16     void init(int n) {
    17         this->n = n;
    18         for (int i = 0; i < n; ++i) G[i].clear();
    19         edges.clear();
    20     }
    21 
    22     void AddEdge(int from, int to, int cap, int cost) {
    23         edges.push_back((Edge){from, to, cap, 0, cost});
    24         edges.push_back((Edge){to, from, 0, 0, -cost});
    25         m = edges.size();
    26         G[from].push_back(m-2); G[to].push_back(m-1);
    27     }
    28 
    29     bool spfa(int s, int t, int &flow, int &cost) {
    30         M(inq, 0); M(d, INF);
    31         d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
    32         queue<int> q;
    33         q.push(s);
    34         while (!q.empty()) {
    35             int x = q.front(); q.pop();
    36             inq[x] = 0;
    37             for (int i = 0; i < G[x].size(); ++i) {
    38                 Edge &e = edges[G[x][i]];
    39                 if (d[e.to] > d[x] + e.cost && e.cap > e.flow) {
    40                     d[e.to] = d[x] + e.cost;
    41                     p[e.to] = G[x][i];
    42                     a[e.to] = min(a[x], e.cap-e.flow);
    43                     if (inq[e.to]) continue;
    44                     q.push(e.to); inq[e.to] = 1;
    45                 }
    46             }
    47         }
    48         if (d[t] == INF) return false;
    49         flow += a[t];
    50         cost += d[t] * a[t];
    51         int u = t;
    52         while (u != s) {
    53             edges[p[u]].flow += a[t];
    54             edges[p[u]^1].flow -= a[t];
    55             u = edges[p[u]].from;
    56         }
    57         return true;
    58     }
    59 
    60     int Mincost(int s, int t) {
    61         int flow = 0, cost = 0;
    62         while (spfa(s, t, flow, cost));
    63         return cost;
    64     }
    65 
    66 };
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  • 原文地址:https://www.cnblogs.com/robin1998/p/6724991.html
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