【BZOJ1076】【SCOI2008】獎勵關（DP、期望、狀壓）

Description

click me

Solution

套路的狀壓期望DP題。。。

考慮倒退期望：設 fi,j f i , j 為一直到第 i−1 i − 1 輪、目前狀态為 j j 的最大分數。

轉移

若目前狀态滿足第kk個寶物的前提條件，那麼選擇取或不取。

若不滿足，那麼不取。

具體轉移方程參看代碼。

Source

/**********************************
 * Au: Hany01
 * Prob: BZOJ1076 & SCOI2008 獎勵關
 * Date: Feb 4th, 2018
 * Email: [email protected]
**********************************/

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

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

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

const int maxn = , maxk = ;

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

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