poj2135 最小费用流

Posted 2020-08-12 王宜鸣

添加超级源点（与点1之间的边容量为2，权值为0）和超级汇点（与点N之间的边容量为2，权值为0），求流量为2的最小费用流。注意是双向边。

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue> 
using namespace std;
const long long INF = 0x3f3f3f3f3f3f3f3f;
typedef long long ll;
typedef pair<ll,int> P;
struct edge
{
    int to,cap;
	ll cost;
	int rev;	
};
int V,E;
vector<edge> G[1005];
ll h[1005];
ll dist[1005];
int prevv[1005];
int preve[1005];
void add_edge(int from,int to,int cap,ll cost)
{
	edge e;
	e.to = to;
	e.cap = cap;
	e.cost = cost;
	e.rev = G[to].size();
	G[from].push_back(e);
	e.to = from;
	e.cap = 0;
	e.cost = -cost;
	e.rev = G[from].size() - 1;
	G[to].push_back(e);
}

ll min_cost_flow(int s,int t,int f)
{
	ll res = 0;
	fill(h,h + V,0);
	while(f > 0)
	{
		priority_queue <P,vector <P>,greater<P> >que;
		fill(dist,dist + V,INF);
		dist[s] = 0;
		que.push(P(0,s));
		while(!que.empty())
		{
			P p = que.top();
			que.pop();
			int v = p.second;
			if(dist[v] < p.first)
			{
				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 + h[v] - h[e.to])
				{
					dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v;
					preve[e.to] = i;
					que.push(P(dist[e.to],e.to));
				}
			}
		}
		if(dist[t] == INF)
		{
			return -1;
		}
		for(int v = 0;v < V;v ++)
		{
			h[v] += dist[v];
		}
		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 * h[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()
{
	int a,b;
	ll c;
    cin >> V >> E;
    
    for(int i = 0;i < E;i ++)
    {
    	scanf("%d%d%lld",&a,&b,&c);
        add_edge(a,b,1,c);
        add_edge(b,a,1,c);
	}
	add_edge(0,1,2,0);
	add_edge(V,V + 1,2,0);
	V += 2;
	cout << min_cost_flow(0,V - 1,2) << endl;
    return 0;	
}