P4015 运输问题 最大/最小费用最大流
Posted wstong
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了P4015 运输问题 最大/最小费用最大流相关的知识,希望对你有一定的参考价值。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 205, inf = 0x3f3f3f3f; 4 struct Edge { 5 int from, to, cap, flow, cost; 6 }; 7 struct MCMF { 8 int n, m, s, t; 9 vector<Edge> edges; 10 vector<int> G[maxn]; 11 int inq[maxn]; 12 int d[maxn]; 13 int p[maxn]; 14 int a[maxn]; 15 16 void init(int n) { 17 this->n = n; 18 for (int i = 1; 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); 27 G[to].push_back(m-1); 28 } 29 bool BellmanFord(int s, int t, int& flow, int& cost) { 30 for (int i = 1; i <= n; ++i) d[i] = inf; 31 memset(inq, 0, sizeof(inq)); 32 d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = inf; 33 34 queue<int> que; 35 que.push(s); 36 while (!que.empty()) { 37 int u = que.front(); que.pop(); 38 inq[u] = 0; 39 for (int i = 0; i < G[u].size(); ++i) { 40 Edge& e = edges[G[u][i]]; 41 if (e.cap > e.flow && d[e.to] > d[u] + e.cost) { 42 d[e.to] = d[u] + e.cost; 43 p[e.to] = G[u][i]; 44 a[e.to] = min(a[u], e.cap-e.flow); 45 if (!inq[e.to]) { que.push(e.to); inq[e.to] = 1; } 46 } 47 } 48 } 49 if (d[t] == inf) return false; 50 flow += a[t]; 51 cost += d[t] * a[t]; 52 int u = t; 53 while (u != s) { 54 edges[p[u]].flow += a[t]; 55 edges[p[u]^1].flow -= a[t]; 56 u = edges[p[u]].from; 57 } 58 return true; 59 } 60 int mincost(int s, int t) { 61 int flow = 0, cost = 0; 62 while (BellmanFord(s, t, flow, cost)); 63 return cost; 64 } 65 }mcmf; 66 int a[maxn], b[maxn], c[maxn][maxn]; 67 int main() { 68 int m, n; scanf("%d%d",&m,&n); 69 int s = m+n+1, t = m+n+2; 70 71 for (int i = 1; i <= m; ++i) scanf("%d",&a[i]); 72 for (int j = 1; j <= n; ++j) scanf("%d",&b[j]); 73 for (int i = 1; i <= m; ++i) { 74 for (int j = 1; j <= n; ++j) { 75 scanf("%d",&c[i][j]); 76 } 77 } 78 // 最小费用最大流 79 mcmf.init(n+m+2); 80 for (int i = 1; i <= m; ++i) { 81 mcmf.AddEdge(s,i,a[i],0); 82 } 83 for (int i = 1; i <= m; ++i) { 84 for (int j = 1; j <= n; ++j) { 85 mcmf.AddEdge(i,j+m,inf,c[i][j]); 86 } 87 } 88 for (int j = 1; j <= n; ++j) { 89 mcmf.AddEdge(j+m,t,b[j],0); 90 } 91 printf("%d ",mcmf.mincost(s,t)); 92 93 // 最大费用最大流 94 mcmf.init(n+m+2); 95 for (int i = 1; i <= m; ++i) { 96 mcmf.AddEdge(s,i,a[i],0); 97 } 98 for (int i = 1; i <= m; ++i) { 99 for (int j = 1; j <= n; ++j) { 100 mcmf.AddEdge(i,j+m,inf,-c[i][j]); 101 } 102 } 103 for (int j = 1; j <= n; ++j) { 104 mcmf.AddEdge(j+m,t,b[j],0); 105 } 106 printf("%d ",-mcmf.mincost(s,t)); 107 return 0; 108 }
以上是关于P4015 运输问题 最大/最小费用最大流的主要内容,如果未能解决你的问题,请参考以下文章