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拆点费用流
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#define rep(i,n) for(int i=0;i<n;++i)
#define clr(x,c) memset(x,c,sizeof(x))
using namespace std;
const int maxn=2000+5;
const int inf=0x3f3f3f3f;
struct Edge {
int from,to,cap,flow,cost;
};
struct mcmf {
bool inq[maxn];
int d[maxn];
int a[maxn];
int p[maxn];
int n,s,t;
vector<int> g[maxn];
vector<Edge> edges;
void init(int n) {
this->n=n;
rep(i,n) g[i].clear();
edges.clear();
}
void addEdge(int u,int v,int cap,int cost) {
edges.push_back( (Edge) {u,v,cap,0,cost} );
edges.push_back( (Edge) {v,u,0,0,-cost} );
g[u].push_back(edges.size()-2);
g[v].push_back(edges.size()-1);
}
bool spfa(int &flow,int &cost) {
clr(d,inf); clr(inq,0);
d[s]=0; inq[s]=1; p[s]=0; a[s]=inf;
queue<int> q;
q.push(s);
while(!q.empty()) {
int x=q.front(); q.pop();
inq[x]=0;
rep(i,g[x].size()) {
Edge &e=edges[g[x][i]];
if(d[e.to]>d[x]+e.cost && e.cap>e.flow) {
d[e.to]=d[x]+e.cost;
p[e.to]=g[x][i];
a[e.to]=min(a[x],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+=d[t]*a[t];
int x=t;
while(x!=s) {
edges[p[x]].flow+=a[t];
edges[p[x]^1].flow-=a[t];
x=edges[p[x]].from;
}
return true;
}
int minCost(int s,int t) {
this->s=s; this->t=t;
int flow=0,cost=0;
while(spfa(flow,cost));
return cost;
}
} g;
int main()
{
// freopen("test.in","r",stdin);
// freopen("test.out","w",stdout);
int n,m;
while(scanf("%d%d",&n,&m)==2) {
g.init(n*2+1);
int a,b,c;
while(m--) {
scanf("%d%d%d",&a,&b,&c);
if(a>1 && a<n) a+=n;
g.addEdge(a,b,1,c);
}
for(int i=2;i<n;++i) g.addEdge(i,i+n,1,0);
g.addEdge(0,1,2,0); g.addEdge(n,n*2,2,0);
printf("%d\n",g.minCost(0,n*2));
}
return 0;
}
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UVa 1658 Admiral(最小费用最大流)
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原文地址：http://www.cnblogs.com/JSZX11556/p/4390906.html
