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POJ-3155-Hard Life(最大密度子图)(01分数规划+最小割)

2017-04-19 21:17 531 查看
链接:http://poj.org/problem?id=3155

求最大密度子图,

见论文:算法合集之《最小割模型在信息学竞赛中的应用》

P20-26

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <string>
#include <map>
#include <stack>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 5007
#define MAXM 5007
int n,m;
const int mod=499999;
#define eps 1e-8
int x[MAXN],y[MAXN],used[MAXN],vis[MAXN],du[MAXN],cnt;
double z[MAXN];
int deep[MAXN];
struct node
{
int v,next;
double flow;
} edge[MAXM];
int ind, head[MAXN];
void add_(int u, int v, double flow)
{
edge[ind].v=v;
edge[ind].flow=flow;
edge[ind].next=head[u];
head[u]=ind++;
edge[ind].v=u;
edge[ind].flow=0;
edge[ind].next=head[v];
head[v]=ind++;
}
bool BFS(int S, int T)
{
memset(deep,0,sizeof(deep));
deep[S]=1;
queue<int> q;
q.push(S);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u]; i+1; i=edge[i].next)
{
int v= edge[i].v;
if(!deep[v]&&edge[i].flow>=eps)
{
deep[v]=deep[u]+1;
q.push(v);
if(v==T)return 1;
}
}
}
return 0;
}
double DFS(int S, int T, double max_)
{
if(S==T)return max_;
double ans=0,f;
for(int i=head[S]; i+1; i=edge[i].next)
{
int v=edge[i].v;
if(deep[S]+1==deep[v] && edge[i].flow>eps)
{
f=DFS(v,T,min(max_-ans,edge[i].flow));
ans+=f;
edge[i].flow-=f;
edge[i^1].flow+=f;
}
}
if(eps<=ans)return ans;
deep[S]=0;
return 0;
}
double Dinic(int S, int T)
{
double ans=0,t;
while(BFS(S,T))
{
while(  ( t=DFS(S,T,INF) ) >=eps)
ans+=t;
}
return ans;
}
void D(int u)
{
vis[u]=1;
if(u>0&&u<=n)cnt++;
for(int i=head[u]; i+1; i=edge[i].next)
{
int v=edge[i].v;
if(!vis[v] && edge[i].flow>eps)D(v);
}
}
void build(double p)
{
ind=0;
memset(head,-1,sizeof(head));
for(int i=0; i<m; ++i)
{
add_(x[i],y[i],1);
add_(y[i],x[i],1);
}
for(int i=1; i<=n; ++i)
{
add_(0,i,m);
add_(i,n+1,m*1.0+2.0*p-du[i]*1.0);
}
}
int main()
{
int cp=0;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(m==0)
{
printf("1\n1\n");
continue ;
}
memset(du,0,sizeof(du));
for(int i=0; i<m; ++i)
{
scanf("%d%d",&x[i],&y[i]);
du[ x[i] ]++;
du[ y[i] ]++;
}
double l=0.0,r=m,ans,mid;
while(r-l>=1.0/n/n)
{
mid=(l+r)/2.0;
ans=0;
build(mid);
double temp=Dinic(0,n+1);
ans=(n*m*1.0-temp)/2.0;
if(ans<eps)r=mid;
else l=mid;
}
cnt=0;
build(l);
Dinic(0,n+1);
memset(vis,0,sizeof(vis));
D(0);
printf("%d\n",cnt);
for(int i=1; i<=n; ++i)
if(vis[i])
printf("%d\n",i);
}
return 0;
}

/*
5 6
1 5
5 4
4 2
2 5
1 2
3 1

5 3
1 3
1 2
2 3

5 5
1 2
1 3
2 3
1 5
1 4

1.5 15
0.75 13.5
1.125 15
0.9375 14.625
1.03125 15
0.984375 14.9062
1.00781 15
*/
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