最小费用最大流模版

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>

using namespace std;

const int MAXN=10100;
const int MAXM=40010;
const int INF=0x3f3f3f3f;

struct Edge      //cost代表单位流量流过该边时,所需的费用大小
{
    int to,next,cap,flow,cost;
}edge[MAXM];

int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;

void init(int n)    //N代表点的个数
{
    N=n;
    tol=0;
    memset(head,-1,sizeof(head));
}

void addEdge(int u,int v,int cap,int cost)    
{
    edge[tol].to=v;
    edge[tol].cap=cap;
    edge[tol].cost=cost;
    edge[tol].flow=0;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].cost=-cost;
    edge[tol].flow=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}

bool spfa(int s,int t)    //计算最短路
{
    queue<int>q;
    for(int i=0;i<N;i++)
    {
        dis[i]=INF;
        vis[i]=false;
        pre[i]=-1;
    }
    dis[s]=0;
    vis[s]=true;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=false;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)
            {
                dis[v]=dis[u]+edge[i].cost;
                pre[v]=i;
                if(!vis[v])
                {
                    vis[v]=true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t]==-1)
        return false;
    else
        return true;
}

int minCostMaxflow(int s,int t,int &cost)    //s代表源点,t代表汇点,cost是最后的总费用,函数返回值是最大流量值,注意点的编号要求从0到N-1
{
    int flow=0;
    cost=0;
    while(spfa(s,t))
    {
        int Min=INF;
        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
        {
            if(Min>edge[i].cap-edge[i].flow)
                Min=edge[i].cap-edge[i].flow;
        }
        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost*Min;
        }
        flow+=Min;
    }
    return flow;
}

 

posted @ 2016-07-30 13:08  相儒以沫  阅读(159)  评论(0编辑  收藏  举报