Topcoder SRM656div1 250 ( 期望DP )

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Problem Statement
    
Charlie has N pancakes. He wants to serve some of them for breakfast. We will number the pancakes 0 through N-1. For each i, pancake i has width i+1 and deliciousness d[i].
Charlie chooses the pancakes he is going to serve using the following randomized process: He starts by choosing the first pancake uniformly at random from all the pancakes he has. He places the chosen pancake onto a plate. This pancake now forms the bottom of a future stack of pancakes. Then, Charlie repeats the following procedure:
If there are no more pancakes remaining, terminate.
Choose a pancake uniformly at random from the pancakes that have not been chosen yet.
If the width of this pancake is greater than the width of the pancake on top of the stack, terminate without taking it.
Place the chosen pancake on top of the stack and go back to step 1.
You are given the vector <int> d with N elements. The total deliciousness of a serving of pancakes is the sum of the deliciousness of all pancakes used in the serving. Compute and return the expected value of the total deliciousness of the pancakes chosen by Charlie.
Definition
    
Class:
RandomPancakeStack
Method:
expectedDeliciousness
Parameters:
vector <int>
Returns:
double
Method signature:
double expectedDeliciousness(vector <int> d)
(be sure your method is public)
Limits
    
Time limit (s):
2.000
Memory limit (MB):
256
Stack limit (MB):
256
Notes
-
Your return value must have an absolute or relative error smaller than or equal to 1e-6
Constraints
-
The number of elements in d will be between 1 and 250, inclusive.
-
Each element of d will be between 1 and 1,000, inclusive.
Examples
0)

    
{1,1,1}
Returns: 1.6666666666666667
The following scenarios may occur:
With probability 1/3, Charlie chooses pancake 0 first. In this case he won‘t be able to add any more pancakes and the total deliciousness of his serving of pancakes will be 1.
With probability 1/3, Charlie chooses pancake 1 first. What happens in the second round? With probability 1/2 he will choose pancake 0 and with probability 1/2 it will be pancake 2. In the first case the total deliciousness of Charlie‘s pancakes will be 2, in the second case it will be 1.
With probability 1/3, Charlie chooses pancake 2 first. If he chooses pancake 0 next, the total deliciousness of his pancakes will be 2. If he happens to choose pancake 1 next (followed by pancake 0 in the third round), the total deliciousness will be 3.
Summing this up, we get the expected deliciousness to be 1/3 * (1) + 1/3 * (1/2 * 1 + 1/2 * 2) + 1/3 * (1/2 * 2 + 1/2 * 3) = 5/3 = 1.666...
1)

    
{3,6,10,9,2}
Returns: 9.891666666666667

2)

    
{10,9,8,7,6,5,4,3,2,1}
Returns: 10.999999724426809

3)

    
{1,2,3,4,5,6,7,8,9,10}
Returns: 7.901100088183421

4)

    
{2,7,1,8,2,8,1,8,2,8,4,5,90,4,5,2,3,5,60,2,8,74,7,1}
Returns: 19.368705050402465

5)

    
{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1}
Returns: 1.718281828459045

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

期望Dp， 不得不承认自己是个概率白痴，这么简单的问题都想了这么久

dp[i][j] 表示选到第i个 ，选了j 个还能选多少个的期望~

#include <bits/stdc++.h>
using namespace std;
const int N = 252 ;
class RandomPancakeStack
{
public:  
    double dp[N][N] , dd[N];
    bool vis[N][N];
    int n ;

    double Dp( int i , int cnt ) {
        if( i == 1 ) return dd[1] ;
        if( !vis[i][cnt] ) {
            vis[i][cnt] = true ;
            double &c = dp[i][cnt] ;
            c = dd[1] ;
            for( int j = 2 ; j <= i ; ++j ) {
                c += 1.0 * ( n - j - cnt ) / ( n - cnt - 1.0 ) * dd[j] + ( 1.0 - 1.0 * ( n - j - cnt ) / ( n - cnt - 1.0 )  ) * ( dd[j] + Dp( j - 1 , cnt + 1 ) ) ;
            }
            c *= 1.0 / i  ;
        }
        return dp[i][cnt] ;
    }

    double expectedDeliciousness( vector <int> d ){
        n = (int) d.size() ;
        for( int i = 0 ; i < n ; ++i ) dd[i+1] = ( double )d[i] ;
        memset( vis , false , sizeof vis );
        return Dp( n , 0 );
    }
};

View Code

Topcoder SRM656div1 250 ( 期望DP )

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原文地址：http://www.cnblogs.com/hlmark/p/4436401.html