【BZOJ1076】【SCOI2008】奖励关（DP、期望、状压）

Description

click me

Solution

套路的状压期望DP题。。。

考虑倒退期望：设fi,jfi,j为一直到第i−1i−1轮、当前状态为jj的最大分数。

转移

若当前状态满足第kk个宝物的前提条件，那么选择取或不取。

若不满足，那么不取。

具体转移方程参看代码。

Source

/**********************************
* Au: Hany01
* Prob: BZOJ1076 & SCOI2008 奖励关
* Date: Feb 4th, 2018
* Email: hany01@foxmail.com
**********************************/

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>

using namespace std;

typedef long long LL;
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define rep(i, k) for (register int i = 0, i##_end_ = (k); i < i##_end_; ++ i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
register int _ = 0, __ = 1; register char c_ = getchar();
for ( ; c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}

inline void File()
{
#ifdef hany01
freopen("bzoj1076.in", "r", stdin);
freopen("bzoj1076.out", "w", stdout);
#endif
}

const int maxn = 16, maxk = 101;

int K, n, p[maxn], pre[maxn], all, tmp;
double f[maxk][1 << maxn];

int main()
{
File();
K = read(), n = read();
For(i, 1, n) {
p[i] = read();
while (tmp = read()) pre[i] |= (1 << (tmp - 1));
}
all = (1 << n);
//f[i][j]: Before the i_th round, when the condition is j, maximize the scores.
Fordown(i, K, 1)
rep(st, all)
{
For(k, 1, n)
if ((pre[k] & st) == pre[k]) f[i][st] += max(f[i + 1][st], f[i + 1][st | (1 << (k - 1))] + p[k]);
else f[i][st] += f[i + 1][st];
f[i][st] /= n;
}
printf("%.6lf\n", f[1][0]);
return 0;
}