P2473 [SCOI2008]奖励关

(P2473) ([SCOI2008])奖励关

令(f[S,i])表示当前吃的状态为(S),当前是第(i)次抛的按最优决策的期望得分

但由于起始状态是确定的,而终态是不确定的,所以考虑改变定义逆推

令(f[S,i])表示(1)到(i-1)的状态为(S),当前是第(i)次抛,还能得到的期望得分

考虑所有抛出的(k)

1. (k)可吃,可吃可不吃 (f[i,S] += max(f[i+1,S],f[i+1,S|(1<<k)]))
2. (k)不可吃 (f[i,S] += f[i+1,S])

弄完后要除(n)

初始值为(0)

答案为(f[1,0])

(tips:) 判断一个集合((S))是否是另一个集合((T))的子集

(S & T == S) (即 (S cap T == S)) 则是

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;

template <typename T> void in(T &x) {
	x = 0; T f = 1; char ch = getchar();
	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
	while(isdigit(ch)) {x = 10*x+ch-'0'; ch = getchar();}
	x *= f;
}

template <typename T> void out(T x) {
	if(x < 0) putchar('-'),x = -x;
	if(x > 9) out(x/10);
	putchar(x%10+'0');
}
//---------------------------------------------------------------

int n,m;
int a[16],need[16];
double f[110][1<<16];

int main() {
	in(m); in(n);
	for(int i = 0;i < n; ++i) {
		in(a[i]); int x;
		while(1) {
			in(x); if(!x) break;
			need[i] |= (1<<(x-1));
		}
	}
	for(int i = m;i >= 1; --i) {
		for(int j = 0;j < (1<<n); ++j) {//debug < (1<<n)  ->  < (1<<n)-1
			for(int k = 0;k < n; ++k) {
				if((j&need[k]) == need[k]) f[i][j] += max(f[i+1][j],f[i+1][j|(1<<k)]+a[k]);
				else f[i][j] += f[i+1][j];
			}
			f[i][j] /= n;
		}
	}
	printf("%.6lf",f[1][0]);
	return 0;
}