卧薪尝胆,厚积薄发。
Hard Life
Date: Fri Sep 07 13:48:13 CST 2018 In Category: NoCategory

Description:

裸最大密度子图。
$1\le n\le 100,1\le m\le 1000$

Solution:

$01$ 分数规划 $+$ 最大权闭合子图。

Code:

不知道为什么 $WA$ 。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 110
#define MAXM 1010
int a[MAXM],b[MAXM];
int s,t;
#define INF 100000000000
struct edge
{
int to,nxt;
long double f;
}e[MAXM * 2 + MAXN * 2 + MAXM * 2 * 2];
int edgenum = 0;
int lin[MAXN + MAXM];
void add(int a,int b,long double k)
{
e[edgenum].to = b;e[edgenum].f = k;e[edgenum].nxt = lin[a];lin[a] = edgenum;++edgenum;
e[edgenum].to = a;e[edgenum].f = 0;e[edgenum].nxt = lin[b];lin[b] = edgenum;++edgenum;
return;
}
int ch[MAXN + MAXM];
#define eps 1e-18
bool BFS()
{
queue<int> q;q.push(s);
memset(ch,-1,sizeof(ch));ch[s] = 0;
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(ch[e[i].to] == -1 && e[i].f >= eps)
{
ch[e[i].to] = ch[k] + 1;
q.push(e[i].to);
}
}
}
return (ch[t] != -1);
}
long double flow(int k,long double f)
{
if(k == t)return f;
long double r = 0;
for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)
{
if(ch[e[i].to] == ch[k] + 1 && e[i].f >= eps)
{
long double l = flow(e[i].to,min(f - r,e[i].f));
e[i].f -= l;r += l;e[i ^ 1].f += l;
}
}
if(r <= eps)ch[k] = -1;
return r;
}
long double dinic()
{
long double ans = 0,r;
while(BFS())while((r = flow(s,1000000000000)) > eps)ans += r;
return ans;
}
bool check(long double x)
{
memset(lin,-1,sizeof(lin));
edgenum = 0;
s = n + m + 1,t = s + 1;
for(int i = 1;i <= m;++i)add(s,i + n,1);
for(int i = 1;i <= n;++i)add(i,t,x);
for(int i = 1;i <= m;++i)
{
add(i + n,a[i],INF);
add(i + n,b[i],INF);
}
long double res = m - dinic();
if(res >= eps)return true;
else return false;
}
vector<int> v;
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1;i <= m;++i)scanf("%d%d",&a[i],&b[i]);
long double l = 0,r = 1000000000000,mid;
long double ans = 0;
while(r - l >= 1e-5)
{
mid = (l + r) / 2;
if(check(mid))
{
ans = mid;
l = mid;
}
else r = mid;
}
check(ans);
for(int i = lin[t];i != -1;i = e[i].nxt)
{
if(e[i ^ 1].f <= eps)v.push_back(e[i].to);
}
sort(v.begin(),v.end());
cout << v.size() << endl;
for(vector<int>::iterator i = v.begin();i != v.end();++i)
{
printf("%d\n",*i);
}
return 0;
}
Copyright © 2020 wjh15101051
ღゝ◡╹)ノ♡