构图方法：
注意题目中的边为无向边。新建源点s 和 汇点t 每两条道路连一条容量为1，费用为w的边。s到1连一条容量为1，费用为0 的边，n到 t 连一条容量为1，费用为0 的边，求最大流。
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#define LL long long
using namespace std;
const int maxn = 10000 + 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) { }
};
int n , m;
vector<Edge>edges;
vector<int>G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init()
{
for(int i=0;i<=n+1;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));
int M = edges.size();
G[from].push_back(M-2);
G[to].push_back(M-1);
}
bool SPFA(int s,int t,int& flow,LL& cost)
{
for(int i=0;i<=n+1;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;
flow += a[t];
cost += (LL)d[t] * (LL)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;
}
int MincostMaxflow(int s,int t,LL &cost)
{
int flow = 0;cost = 0;
while(SPFA(s,t,flow,cost));
return flow;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
int u , v , w;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
AddEdge(u,v,1,w);
AddEdge(v,u,1,w);
}
int s = 0 , t = n + 1;
AddEdge(s,1,2,0);
AddEdge(n,t,2,0);
LL cost = 0;
int ans = MincostMaxflow(s,t,cost);
printf("%lld\n",cost);
}
return 0;
}
POJ 2135 Farm Tour (dinic算法，网络流),布布扣,bubuko.com
POJ 2135 Farm Tour (dinic算法，网络流)
原文地址：http://blog.csdn.net/moguxiaozhe/article/details/38403703
