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  • 最小费用最大流模板

    新的模板


    #include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    const int OO=1e9;//无穷大
    const int maxm=1111111;//边的最大数量,为原图的两倍
    const int maxn=2222;//点的最大数量
    
    int node,src,dest,edge;//node节点数,src源点,dest汇点,edge边数
    int head[maxn],p[maxn],dis[maxn],q[maxn],vis[maxn];//head链表头,p记录可行流上节点对应的反向边,dis计算距离
    
    struct edgenode
    {
        int to;//边的指向
        int flow;//边的容量
        int cost;//边的费用
        int next;//链表的下一条边
    } edges[maxm];
    
    void prepare(int _node,int _src,int _dest);
    void addedge(int u,int v,int f,int c);
    bool spfa();


    inline int min(int a,int b)
    {
        return a<b?a:b;
    }
    
    inline void prepare(int _node,int _src,int _dest)
    {
        node=_node;
        src=_src;
        dest=_dest;
        for (int i=0; i<node; i++)
        {
            head[i]=-1;
            vis[i]=false;
        }
        edge=0;
    }
    
    void addedge(int u,int v,int f,int c)
    {
        edges[edge].flow=f;
        edges[edge].cost=c;
        edges[edge].to=v;
        edges[edge].next=head[u];
        head[u]=edge++;
        edges[edge].flow=0;
        edges[edge].cost=-c;
        edges[edge].to=u;
        edges[edge].next=head[v];
        head[v]=edge++;
    }
    
    bool spfa()
    {
        int i,u,v,l,r=0,tmp;
        for (i=0; i<node; i++) dis[i]=OO;
        dis[q[r++]=src]=0;
        p[src]=p[dest]=-1;
        for (l=0; l!=r; ((++l>=maxn)?l=0:1))
        {
            for (i=head[u=q[l]],vis[u]=false; i!=-1; i=edges[i].next)
            {
                if (edges[i].flow&&dis[v=edges[i].to]>(tmp=dis[u]+edges[i].cost))
                {
                    dis[v]=tmp;
                    p[v]=i^1;
                    if (vis[v]) continue;
                    vis[q[r++]=v]=true;
                    if (r>=maxn) r=0;
                }
            }
        }
        return p[dest]>=0;
    }
    
    int spfaflow()
    {
        int i,ret=0,delta;
        while (spfa())
        {
            //按记录原路返回求流量
    
            for (i=p[dest],delta=OO; i>=0; i=p[edges[i].to])
            {
                delta=min(delta,edges[i^1].flow);
            }
            for (int i=p[dest]; i>=0; i=p[edges[i].to])
            {
                edges[i].flow+=delta;
                edges[i^1].flow-=delta;
            }
            ret+=delta*dis[dest];
        }
        return ret;
    }
    


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  • 原文地址:https://www.cnblogs.com/cyendra/p/3038443.html
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