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\n1\n");

        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\n",num);

    for(i=0;i<num;i++)

        printf("%d\n",ans[i]);

    }

    return 0;

}