poj 3155 最大密度子图

思路：

这个还是看的胡伯涛的论文《最小割在信息学竞赛中的应用》。是将最大密度子图问题转化为了01分数规划和最小割问题。

直接上代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#define Maxn 6010
#define Maxm 200000
#define LL double
#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-5;
LL value[Maxn];
int head[Maxn],work[Maxn],dis[Maxn],q[Maxn],e,vi[Maxn];
inline void addedge(int from,int to,LL val)//有向边
{
    edge[e].from=from;
    edge[e].to=to;
    edge[e].val=val;
    edge[e].next=head[from];
    head[from]=e++;
    edge[e].from=to;
    edge[e].to=from;
    edge[e].val=0;
    edge[e].next=head[to];
    head[to]=e++;
}
inline double min(double a,double b)
{
    return a>b?b:a;
}
void init()
{
    e=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,LL c)
{
    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++;
}
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>0&&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>0&&dis[edge[i].to]==dis[cur]+1)
        {
            LL t=dfs(edge[i].to,min(a,edge[i].val),T);
            if(t>0)
            {
                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));
        LL t=dfs(S,inf,T);
        while(t>0)
        {
            ans+=t;
            t=dfs(S,inf,T);
        }
    }
    return ans;
}

int main()
{
    int n,m,i,j,a[Maxn],b[Maxn];
    int degree[Maxn];
    memset(degree,0,sizeof(degree));
    while(scanf("%d%d",&n,&m)!=EOF)
    {
    if(m==0)
    {
        printf("1
1
");
        return 0;
    }
    init();
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",a+i,b+i);
        degree[a[i]]++;
        degree[b[i]]++;
    }
    double l=0,r=m,mid;
    double eps2=1.0/n/n;
    while(r-l>eps2)
    {
        mid=(l+r)/2;
        init();
        for(i=1;i<=m;i++)
        {
            add(a[i],b[i],1);
            add(b[i],a[i],1);
        }
        for(i=1;i<=n;i++)
        {
            add(0,i,m);
            add(i,n+1,m * 1.0 + 2 * mid - degree[i] * 1.0);

        }
        double tt=Dinic(0,n+1);
        double temp=(m*n*1.0-tt)/2.0;
        //cout<<tt<<endl;
        if(temp>eps)
            l=mid;
        else
            r=mid;
    }
    init();
    for(i=1;i<=m;i++)
    {
        add(a[i],b[i],1);
        add(b[i],a[i],1);
    }
    for(i=1;i<=n;i++)
    {
        add(0,i,m);
        add(i,n+1,m*1.0+2*l-degree[i]*1.0);
        //cout<<m<<" "<<mid<<" "<<degree[i]<<" "<<m+2*mid-degree[i]<<endl;
    }
    //for(i=0;i<e;i++)
        //cout<<edge[i].to<<" "<<edge[i].next<<" "<<edge[i].val<<endl;
    Dinic(0,n+1);
    vector<int> ans;
    memset(vi,0,sizeof(vi));
    for(i=1;i<=n;i++)
        if(dis[i]>=0)
            ans.push_back(i);
    int num=ans.size();
    printf("%d
",num);
    for(i=0;i<num;i++)
        printf("%d
",ans[i]);
    }
    return 0;
}