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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;
}
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原文地址:http://www.cnblogs.com/wangyiming/p/5929679.html