bzoj4036(min-max容斥+FMT)

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=4036

题解

裸 min-max 容斥,题目求的是 $\max {U}$ ,可以转化为求每个集合的 $\min{T}$ ,然后显然有

然后对每个集合枚举元素即可,这里用了 FMT ,复杂度同样为 $O(n2^n)$




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
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*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
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*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 1048576
#define nm 1024
#define pi 3.1415926535897931
using namespace std;
const ll inf=998244353;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}





int n,m;
double a[NM],ans;

void fmt(double*a){
for(int i=0;succ(i)<n;i++)
inc(j,0,n-1)if(j>>i&1)
a[j]=a[j]+a[j^1<<i];
}

int main(){
m=read();
n=succ(m);
inc(i,0,n-1)scanf("%lf",a+i);
fmt(a);
inc(i,0,n-2){
if(fabs(a[i]-1)<eps)return 0*printf("INF\n");
if((m-__builtin_popcount(i)+1)&1)ans-=1/(1-a[i]);
else ans+=1/(1-a[i]);
}
return 0*printf("%.7lf\n",ans);
}