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一道最大密度子图的模板题

一道卡精度的神题

二分精度不能太大，网络流精度不能太小 （：

//Achen
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<cstdio>
#include<queue>
#include<cmath>
const int N=12007;
typedef double LL;
const LL inf=1e18,eps=1e-12; 
using namespace std;
int n,m,num;

template<typename T>void read(T &x)  {
    char ch=getchar(); x=0; T f=1;
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') f=-1,ch=getchar();
    for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
}

struct edge{
    int from,to,nx;
    LL cap,flow;
    edge(){}
    edge(int from,int to,LL cap,LL flow,int nx):from(from),to(to),cap(cap),flow(flow),nx(nx){}
}e[N];

int ecnt=1,fir[N],dd[N];
void add(int u,int v,LL cap) {
    e[++ecnt]=edge(u,v,cap,0,fir[u]); fir[u]=ecnt;
    e[++ecnt]=edge(v,u,0,0,fir[v]); fir[v]=ecnt;
}

queue<int>que;
int c[N],d[N],cur[N],p[N];
void bfs(int s,int t) {
    for(int i=0;i<=num;i++) d[i]=num,c[i]=0;
    que.push(t); d[t]=0;
    while(!que.empty()) {
        int x=que.front();
        que.pop();
        for(int i=fir[x];i;i=e[i].nx) {
            int y=e[i].to;
            if(d[y]==num&&e[i].cap==0) {
                d[y]=d[x]+1;
                que.push(y);
            }
        }
    } 
}

LL cal(int s,int t) {
    LL fl=inf;
    for(int i=t;i!=s;i=e[p[i]].from)
        fl=min(fl,e[p[i]].cap-e[p[i]].flow);
    for(int i=t;i!=s;i=e[p[i]].from) {
        if(p[i]==886||((p[i]^1)==886)) {
            int debug=1;
        }
        e[p[i]].flow=e[p[i]].flow+fl,
        e[p[i]^1].flow-=fl;
    }
    return fl;
}

LL ISAP(int s,int t) {
    bfs(s,t);
    LL res=0;
    for(int i=1;i<=num;i++) c[d[i]]++,cur[i]=fir[i];
    for(int x=s;d[x]<num;) {
        if(x==t) {
            res+=cal(s,t);
            x=s;
        } 
        int ok=0;
        for(int &i=cur[x];i;i=e[i].nx) 
            if(e[i].flow+eps<=e[i].cap&&d[e[i].to]+1==d[x]) {
                ok=1; p[x=e[i].to]=i; break;
            }
        if(!ok) {
            cur[x]=fir[x]; int M=num;
            for(int i=fir[x];i;i=e[i].nx) 
                if(e[i].flow+eps<=e[i].cap) M=min(M,d[e[i].to]+1);
            if(!(--c[d[x]])) break;
            c[d[x]=M]++;
            if(x!=s) x=e[p[x]].from;
        }
    }
    return res;
}

int vis[N],ansnum; 
void dfs(int x) {
    for(int i=fir[x];i;i=e[i].nx) 
        if(!vis[e[i].to]&&e[i].cap-e[i].flow>=eps) { 
            vis[e[i].to]=1;
            ansnum++;
            dfs(e[i].to);
        }
}

int ee[N][2];

int ck(LL mid,int o) {
    int s=n+1,t=n+2,res=0; num=t;
    ecnt=1;
    memset(fir,0,sizeof(fir));    
    for(int i=1;i<=m;i++) {
        add(ee[i][0],ee[i][1],1);
        add(ee[i][1],ee[i][0],1);
    }
    for(int i=1;i<=n;i++) {
        add(s,i,(LL)m);
        add(i,t,(LL)m+2.0*mid-dd[i]);
    }
    if(((LL)n*m-ISAP(s,t))/2.0>=eps) res=1;
    if(o==1) {
        memset(vis,0,sizeof(vis));  
        vis[s]=1; dfs(s);
    }
    return res;
}

int main() {
    while(scanf("%d%d",&n,&m)==2) {
        if( m == 0 ) {  
            printf("1
1
") ;  
            continue ;  
        }
        memset(dd,0,sizeof(dd));
        for(int i=1;i<=m;i++) {
            read(ee[i][0]); read(ee[i][1]);
            dd[ee[i][0]]++; dd[ee[i][1]]++;
        }
        LL l=0,r=m;
        while(l+1e-5<=r) {
            LL mid=(l+r)/2.0;
            if(ck(mid,0)) l=mid;
            else r=mid;
        }
        ansnum=0;
        ck(l,1); 
        printf("%d
",ansnum);
        for(int i=1;i<=n;i++) 
            if(vis[i]) printf("%d
",i);
    }
    return 0;
}