14西安区域赛C - The Problem Needs 3D Arrays

最大密度子图裸题，详情请见胡博涛论文：
https://wenku.baidu.com/view/986baf00b52acfc789ebc9a9.html
不加当前弧优化t到死= =

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,  b) ((a)>(b)?(a):(b))
#define Min(a,  b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const int N=20000+10,maxn=50000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

struct edge{
    int to,Next;
    double c;
}e[maxn];
int s,t,cnt,head[N],cur[N];
void init()
{
    cnt=0;
    memset(head,-1,sizeof head);
}
void add(int u,int v,double c)
{
    e[cnt].to=v;
    e[cnt].c=c;
    e[cnt].Next=head[u];
    head[u]=cnt++;
    e[cnt].to=u;
    e[cnt].c=0;
    e[cnt].Next=head[v];
    head[v]=cnt++;
}
inline bool zero(double x){return fabs(x)<=eps;}
int dis[N];
bool bfs()
{
    queue<int>q;
    memset(dis,-1,sizeof dis);
    dis[s]=1;
    q.push(s);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=head[x];~i;i=e[i].Next)
        {
            int y=e[i].to;
            if(dis[y]==-1&&!zero(e[i].c))
            {
                dis[y]=dis[x]+1;
                q.push(y);
            }
        }
    }
    return dis[t]!=-1;
}
double dfs(int u,double mx)
{
    if(u==t)return mx;
    double flow=0,f;
    for(int &i=cur[u];~i;i=e[i].Next)
    {
        int x=e[i].to;
        if(dis[x]==dis[u]+1&&!zero(e[i].c)&&!zero(f=dfs(x,min(mx-flow,e[i].c))))
        {
            e[i].c-=f;
            e[i^1].c+=f;
            return f;
        }
    }
    return 0;
}
double maxflow()
{
    double ans=0,f;
    while(bfs())
    {
        for(int i=0;i<=t;i++)cur[i]=head[i];
        while(!zero(f=dfs(s,inf)))ans+=f;
    }
    return ans;
}
int a[N],deg[N],n,m;
pii p[maxn];
bool ok(double x)
{
     init();
     s=n+m+1,t=n+m+2;
     for(int i=1;i<=n;i++)add(i,t,x);
     for(int i=1;i<=m;i++)
     {
         add(s,n+i,1);
         add(n+i,p[i].fi,inf);
         add(n+i,p[i].se,inf);
     }
     return !zero(m-maxflow());
}
double solve()
{
    double l=0,r=m;
    while(r-l>=1e-8)
    {
        double mid=(l+r)/2;
        if(ok(mid))l=mid;
        else r=mid;
    }
    return r+eps;
}

int main()
{
    int T;scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        memset(deg,0,sizeof deg);
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        m = 0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(a[j]<a[i])
                {
                    p[++m]=mp(i,j);
                    deg[i]++,deg[j]++;
                }
            }
        }
        printf("Case #%d: %.8f\n",cas,solve());
    }
    return 0;
}
/********************
2
5
3 4 2 5 1
5
3 4 2 5 1
********************/