本文主要是介绍[POJ2047] Concert Hall Scheduling 费用流，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
在SPFA的双端队列优化的时候打错了.. 把d[v] > d[q.front()] 打成了 d[v] > q.front() 结果瞬间快十倍直接AC了= =
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<deque>
#define SF scanf
#define PF printf
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 2000;
struct Node { int u, v, c, f, cost;
};
struct node {int s, t, w;bool operator < (const node &t) const {return s < t.s;}
}A[MAXN+10];
template <int maxn>
struct MCMF { int n, m, s, t; vector <Node> Edge; vector <int> G[maxn+10]; int inq[maxn+10], d[maxn+10], p[maxn+10], a[maxn+10]; void init(int n) { this->n = n; for(int i = 0; i < n; i++) G[i].clear(); Edge.clear(); } void add(int u, int v, int c, int cost) { Edge.push_back((Node) { u, v, c, 0, cost } ); Edge.push_back((Node) { v, u, 0, 0, -cost } ); m = Edge.size(); G[u].push_back(m-2); G[v].push_back(m-1); } bool SPFA(int s, int t, int &Flow, int &Cost) { for(int i = 0; i <= t; i++) d[i] = -INF;memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF; deque <int> q; q.push_front(s); while(!q.empty()) { int u = q.front(); q.pop_front(); inq[u] = 0; for(int i = 0; i < G[u].size(); i++) { int v = Edge[G[u][i]].v, c = Edge[G[u][i]].c, f = Edge[G[u][i]].f, cost = Edge[G[u][i]].cost; if(c > f && d[v] < d[u] + cost) { d[v] = d[u] + cost; p[v] = G[u][i]; a[v] = min(a[u], c - f); if(!inq[v]) {inq[v] = true; if(d[v] > q.front()) q.push_front(v);else q.push_back(v);} } } } if(d[t] == -INF) return false; Flow += a[t]; Cost += d[t] * a[t]; int u = t; while(u != s) { Edge[p[u]].f += a[t]; Edge[p[u]^1].f -= a[t]; u = Edge[p[u]].u; } return true; } int Mincost(int s, int t) { int Flow = 0, Cost = 0; for(int i = 0; i < 2; i++)SPFA(s, t, Flow, Cost); return Cost; }
};
MCMF <MAXN+10> sap;
int main()
{int n;while(SF("%d", &n) && n){for(int i = 1; i <= n; i++) SF("%d%d%d", &A[i].s, &A[i].t, &A[i].w);sort(A+1, A+1+n);int S = 0, T = 2 * n + 1;sap.init(T);for(int i = 1; i <= n; i++){sap.add(S, i, 1, 0);sap.add(i, i+n, 1, A[i].w);sap.add(i+n, T, 1, 0);}for(int i = 1; i <= n; i++){int j = i + 1;while(A[j].s <= A[i].t) j++;for( ; j <= n; j++) sap.add(i+n, j, 1, 0);}int ans = sap.Mincost(S, T);PF("%d\n", ans);}
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