BZOJ 1076: [SCOI2008]奖励关

期望dp一般逆推，因为终点不确定，而起点确定
那么这道题就可以定义状态
\(f_{i,j}\) 为第 \(i\) 次抛出宝物到结束种类集合为 \(j\) 的期望收益
则答案为 \(f_{0,0}\)
而dp过程就可以从后往前叠加贡献
复杂度 \(O(kn2^n)\)

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define ll long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int cnt=1,head[N],to[N*2],ne[N*2];void addd(int u,int v){to[++cnt]=v;ne[cnt]=head[u];head[u]=cnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int cnt=1,head[N],to[N*2],ne[N*2],c[N*2];void addd(int u,int v,int w){to[++cnt]=v;ne[cnt]=head[u];c[cnt]=w;head[u]=cnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 107;
const int M = 1 << 15 | 10;
db f[N][M];
int score[N], need[N], k, n;

int main() {
	k = _(), n = _();
	rep(i, 0, n) {
		score[i] = _();
		int t = _();
		while (t) {
			need[i] |= 1 << t - 1;
			t = _();
		}
	}
	per(i, 0, k) {
		rep(j, 0, (1 << n)) {
			rep(z, 0, n) {
				if ((j & need[z]) == need[z])
					f[i][j] += std::max(f[i + 1][j], f[i + 1][j | (1 << z)] + score[z]);
				else
					f[i][j] += f[i + 1][j];
			}
			f[i][j] /= (db)n;
		}
	}
	printf("%.6f\n", f[0][0]);
	return 0;
}