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  • 高效的网络流dinic算法模版

    #include <cstring>
    #include <algorithm>
    #include <vector>
    #define Maxn 120010
    #define Maxm 210000
    #define LL int
    #define inf 100000000
    #define Abs(a) (a)>0?(a):(-a)
    using namespace std;
    struct Edge{
        int from,to,next;
        LL val;
    }edge[Maxm];
    const double eps=1e-9;
    LL value[Maxn];
    int head[Maxn],work[Maxn],dis[Maxn],q[Maxn],e,vi[Maxn];
    void init()
    {
        e=0;
        memset(head,-1,sizeof(head));
    }
    void add1(int u,int v,LL c)//有向边
    {
        edge[e].to=v;edge[e].val=c;edge[e].next=head[u];head[u]=e++;
        edge[e].to=u;edge[e].val=0;edge[e].next=head[v];head[v]=e++;
    }
    void add2(int u,int v,LL c)//无向边
    {
        edge[e].to=v;edge[e].val=c;edge[e].next=head[u];head[u]=e++;
        edge[e].to=u;edge[e].val=c;edge[e].next=head[v];head[v]=e++;
    }
    int bfs(int S,int T)
    {
        int rear=0;
        memset(dis,-1,sizeof(dis));
        dis[S]=0;q[rear++]=S;
        for(int i=0;i<rear;i++)
        {
            for(int j=head[q[i]];j!=-1;j=edge[j].next)
            {
                if(edge[j].val&&dis[edge[j].to]==-1)
                {
                    dis[edge[j].to]=dis[q[i]]+1;
                    q[rear++]=edge[j].to;
                    if(edge[j].to==T) return 1;
                }
            }
        }
        return 0;
    }
    LL dfs(int cur,LL a,int T)
    {
        if(cur==T) return a;
        for(int &i=work[cur];i!=-1;i=edge[i].next)
        {
            if(edge[i].val&&dis[edge[i].to]==dis[cur]+1)
            {
                LL t=dfs(edge[i].to,min(a,edge[i].val),T);
                if(t)
                {
                    edge[i].val-=t;
                    edge[i^1].val+=t;
                    return t;
                }
            }
        }
        return 0;
    }
    LL Dinic(int S,int T)
    {
        LL ans=0;
        while(bfs(S,T))
        {
            memcpy(work,head,sizeof(head));
            while(LL t=dfs(S,inf,T)) ans+=t;
        }
        return ans;
    }
    int main()
    {
        int n,m,i,j,num=0;
        Dinic(0,n+1);
        for(i=1;i<=n;i++)//找出S的点的个数
            if(dis[i]>=0)
                num++;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wangfang20/p/3219169.html
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