BZOJ4481 JSOI2015非诚勿扰（概率期望+树状数组）

　　首先求出每个女性接受某个男性的概率。这个概率显然是一个无穷等比数列求和。

　　然后按编号从小到大考虑每个女性，维护出每个男性被选择的期望次数，BIT上查询后缀和即可。

　　需要long double。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 500010
#define double long double
int n,m;
double p,tree[N],q[N],ans;
vector<int> a[N];
double query(int k){double s=0;while (k) s+=tree[k],k-=k&-k;return s;}
void ins(int k,double s){while (k<=n) tree[k]+=s,k+=k&-k;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4481.in","r",stdin);
    freopen("bzoj4481.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();cin>>p;
    for (int i=1;i<=m;i++)
    {
        int x=read(),y=read();
        a[x].push_back(y);
    }
    int cnt=0;
    for (int i=1;i<=n;i++)
    {
        sort(a[i].begin(),a[i].end());
        int s=a[i].size();
        double v=1;
        for (int j=0;j<s;j++) q[j]=v*p,v*=1-p;
        for (int j=0;j<s;j++) q[j]/=1-v;
        for (int j=0;j<s;j++) ans+=q[j]*(cnt-query(a[i][j]));
        for (int j=0;j<s;j++) ins(a[i][j],q[j]);
        cnt+=s>0;
    }
#undef double
    double p=ans;printf("%.2f",p);
    return 0;
}