hdu4336--expectation + State compression Dp--card Collector

http://acm.hdu.edu.cn/showproblem.php?pid=4336

Transfer from http://www.cnblogs.com/zhj5chengfeng/archive/2013/03/02/2939601.html

Approach Analysis

Since there are only 20 cards, using state compression, 0 means the card is not collected, and 1 means the card is collected

Order: F[s] Represents the status of the card type that has been set, the expectation that all cards need to buy things

Buy a thing, the bag may:

1. No card

2. The card is already collected

3. The card is not collected

So there are:

F[s] = 1 + ((1-segma{p[i}) F[s]) + (segma{P[j]*f[s]}) + (segma{p[k]*f[s| ( 1<<k)]})

Including: i=0,2,..., n-1

J= J cards have been collected, i.e. s from right to left number J is 1:s& (1<<J)!=0

K= k cards are not collected, i.e. s from right to left number K bit is 0:s& (1<<k) ==0

The move item can be:

segma{P[i]}f[s] = 1 + segma{p[i]*f[s| ( 1<<i)},i= I card not collected

Target status is: F[0]

/************************************************* Author:P owatr* Created time:2015-8-25 18:21:57* File Name : Hdu4336.cpp ************************************************/#include <cstdio> #include <algorithm># Include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string > #include <vector> #include <queue> #include <deque> #include <stack> #include <list># Include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime>using namespace std; #define Lson L, Mid, RT << 1#define Rson mid + 1, R, RT << 1 | 1typedef long ll;const int MAXN = 1e5 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;double dp[1<<21];    Double A[21];int Main () {int n;        while (~SCANF ("%d", &n)) {for (int i = 0; i < n; i++) scanf ("%lf", &a[i]);        memset (DP, 0, sizeof (DP)); int MAXN = (1 << N)-1;       DP[MAXN] = 0;            for (int i = maxn-1; I >= 0; i--) {double sum = 0;            Dp[i] = 1;                for (int j = 0; J < N; j + +) {if (I & (1 << j)) continue; Dp[i] + = dp[i| (                1&LT;&LT;J)]*a[j];            Sum + = A[j];        } Dp[i]/= sum;    } printf ("%f\n", dp[0]);             } return 0;}

　　

hdu4336--expectation + State compression Dp--card Collector