CF739E Gosha is hunting 【WQS二分 + 期望】

题目链接

CF739E

题解

抓住个数的期望即为概率之和

使用\(A\)的期望为\(p[i]\)

使用\(B\)的期望为\(u[i]\)

都使用的期望为\(p[i] + u[i] - u[i]p[i]\)

当然是用越多越好

但是他很烦地给了个上限，我们就需要作出选择了

有一个很明显的\(O(n^3)\)的\(dp\)，显然过不了

但我们有一个很好的\(WQS\)二分

我们非常想去掉这个上限

那就去掉吧，但是每用一次都要付出一个代价

我们二分这个代价，当使用次数恰好为为\(a\)和\(b\)时就是答案

再加回付出的代价即可

非常巧妙地变成了\(O(n\log^2n)\)

这种二分技巧非常棒

当我们求的东西有一个限制个数时，可以通过设置代价去掉上限

//Mychael
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
#define eps 1e-9
using namespace std;
const int maxn = 2005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
	return flag ? out : -out;
}
int n,a,b,cnta,cntb;
double p[maxn],u[maxn],A,B,ans;
int work(double cost){
	A = cost; cnta = cntb = 0; ans = 0;
	int sol; double val;
	REP(i,n){
		val = 0; sol = 0;
		if (p[i] - A > val) sol = 1,val = p[i] - A;
		if (u[i] - B > val) sol = 2,val = u[i] - B;
		if (p[i] + u[i] - u[i] * p[i] - A - B > val)
			sol = 3,val = p[i] + u[i] - u[i] * p[i] - A - B;
		if (sol == 1 || sol == 3) cnta++;
		if (sol == 2 || sol == 3) cntb++;
		ans += val;
	}
	return cnta;
}
int check(double cost){
	B = cost;
	double l = 0,r = 1.0,mid;
	while (r - l > eps){
		mid = (l + r) / 2.0;
		if (work(mid) <= a) r = mid;
		else l = mid;
	}
	work(r);
	A = l;
	return cntb;
}
int main(){
	n = read(); a = read(); b = read();
	REP(i,n) scanf("%lf",&p[i]);
	REP(i,n) scanf("%lf",&u[i]);
	double l = 0,r = 1.0,mid;
	while (r - l > eps){
		mid = (l + r) / 2.0;
		if (check(mid) <= b) r = mid;
		else l = mid;
	}
	check(r);
	printf("%.8lf",ans + a * A + b * B);
	return 0;
}