aoj GRL_6 B Network Flow - Minimum Cost Flow
問題概要
省略
解法
最小費用流
ミス
蟻本のやつを写経した。
コード
#include <iostream> #include <cstdio> #include <vector> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; #define rep(i,n) for(int i=0;i<(n);i++) const int INF = 1e9; struct edge { int to, cap, cost, rev; }; int V; const int MAX_V = 100; vector<edge> G[MAX_V]; int dist[MAX_V]; int prevv[MAX_V], preve[MAX_V]; void add_edge(int from, int to, int cap, int cost) { G[from].push_back((edge){to, cap, cost, (int)G[to].size()}); G[to].push_back((edge){from, 0, -cost, (int)G[from].size() - 1}); } int min_cost_flow(int s, int t, int f) { int res = 0; while (f > 0) { fill(dist, dist + V, INF); dist[s] = 0; bool update = true; while (update) { update = false; for (int v = 0; v < V; v++) { if (dist[v] == INF) continue; for (int i = 0; i < G[v].size(); i++) { edge &e = G[v][i]; if (e.cap > 0 && dist[e.to] > dist[v] + e.cost) { dist[e.to] = dist[v] + e.cost; prevv[e.to] = v; preve[e.to] = i; update = true; } } } } if (dist[t] == INF) { return -1; } int d = f; for (int v = t; v != s; v = prevv[v]) { d = min(d, G[prevv[v]][preve[v]].cap); } f -= d; res += d * dist[t]; for (int v = t; v != s; v = prevv[v]) { edge &e = G[prevv[v]][preve[v]]; e.cap -= d; G[v][e.rev].cap += d; } } return res; } int main(void){ cin >> V; int e, f; cin >> e >> f; rep(i, e){ int u, v, c, d; cin >> u >> v >> c >> d; add_edge(u, v, c, d); } printf("%d\n", min_cost_flow(0, V - 1, f)); }