uva1658 Admiral
拆点法是很套路的方法,将一个点拆为i和i',连边,cap为1,cost为0.
编号:点2~n-1拆成弧i->i',前者编号为0~n-1,后者编号为n~2n-3 (i+n-2)
本题拆点后从1-v求流量为2的最小费用流
while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
注意和没有上限的最小费用最大流代码的区别
#include<cstdio> #include<cstring> #include<queue> #include<vector> #include<algorithm> #include<cassert> using namespace std; const int maxn = 2000 + 10; const int INF = 1000000000; struct Edge { int from, to, cap, flow, cost; Edge(int u, int v, int c, int f, int w):from(u),to(v),cap(c),flow(f),cost(w) {} }; struct MCMF { int n, m; vector<Edge> edges; vector<int> G[maxn]; int inq[maxn]; // 是否在队列中 int d[maxn]; // Bellman-Ford int p[maxn]; // 上一条弧 int a[maxn]; // 可改进量 void init(int n) { this->n = n; for(int i = 0; i < n; i++) G[i].clear(); edges.clear(); } void AddEdge(int from, int to, int cap, int cost) { edges.push_back(Edge(from, to, cap, 0, cost)); edges.push_back(Edge(to, from, 0, 0, -cost)); m = edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost) { for(int i = 0; i < n; i++) d[i] = INF; memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF; queue<int> Q; Q.push(s); while(!Q.empty()) { int u = Q.front(); Q.pop(); inq[u] = 0; for(int i = 0; i < G[u].size(); i++) { Edge& e = edges[G[u][i]]; if(e.cap > e.flow && d[e.to] > d[u] + e.cost) { d[e.to] = d[u] + e.cost; p[e.to] = G[u][i]; a[e.to] = min(a[u], e.cap - e.flow); if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; } } } } if(d[t] == INF) return false; if(flow + a[t] > flow_limit) a[t] = flow_limit - flow; flow += a[t]; cost += d[t] * a[t]; for(int u = t; u != s; u = edges[p[u]].from) { edges[p[u]].flow += a[t]; edges[p[u]^1].flow -= a[t]; } return true; } // 需要保证初始网络中没有负权圈 int MincostFlow(int s, int t, int flow_limit, int& cost) { int flow = 0; cost = 0; while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost)); return flow; } }; MCMF g; int main() { int n, m, a, b, c; while(scanf("%d%d", &n, &m) == 2 && n) { g.init(n*2-2); // 点2~n-1拆成弧i->i',前者编号为0~n-1,后者编号为n~2n-3 for(int i = 2; i <= n-1; i++) g.AddEdge(i-1, i+n-2, 1, 0); while(m--) { scanf("%d%d%d", &a, &b, &c); // 连a'->b if(a != 1 && a != n) a += n-2; else a--; b--; g.AddEdge(a, b, 1, c); } int cost; g.MincostFlow(0, n-1, 2, cost); printf("%d\n", cost); } return 0; }