ZOJ 3551 —— 概率DP

Bloodsucker                                       Time Limit: 2 Seconds       Memory Limit: 65536 KB In 0th day, there are n-1 people and 1 bloodsucker. Every day, two and only two of them meet.  Nothing will happen if they are of the same species,  that is, a people meets a people or a bloodsucker meets a bloodsucker. Otherwise, people may  be transformed into bloodsucker with probability p. Sooner or later(D days), all people will be turned into bloodsucker. Calculate the mathematical expectation of D. Input The number of test cases (T, T ≤ 100) is given in the first line of the input. Each case consists of an integer n and a float number p (1 ≤ n < 100000, 0 < p ≤ 1, accurate to 3 digits after decimal point),  separated by spaces. Output For each case, you should output the expectation(3 digits after the decimal point) in a single line. Sample Input 1 2 1 Sample Output 1.000 Author:  WU, Yingxin Contest:  ZOJ Monthly, October 2011 Submit     Status

题意是有n - 1个人和1个吸血鬼，每天都会有两个人相遇，如果一个是人一个是吸血鬼，

那么人有p的概率变成吸血鬼， 问你所有人都变成吸血鬼的期望天数。

思路：有两种解法:

一、利用公式期望等于概率分之一，直接扫一遍求和即可。

       请见http://blog.chinaunix.net/uid-26372629-id-2996052.html

二、我们用dp[i]表示从i个吸血鬼到n个吸血鬼的天数期望。

       pa表示人与吸血鬼相遇的概率。C（1 ， i） * C（1 ， n-i） / C（2 ， n）

       pp表示相遇且成功变成吸血鬼的概率。pp = pa * p

       dp[i] = (dp[i + 1] + 1) * pp + (dp[i] + 1) * (1 - pp)            这里的+1表示加了一天

       可以简化成dp[i] = (dp[i + 1] * pp + 1) / pp

       逆推一遍即可得到答案。

/*
ID: xinming2
PROG: stall4
LANG: C++
*/
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 100000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int mod = (int)1e9 + 7;
typedef long long LL;
const LL MOD = 1000000007LL;
const double PI = acos(-1.0);

typedef pair<int , int> pi;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
double dp[MAXN];
int main()
{
    int n;
    double p;
    int T;
    scanf("%d" , &T);
    while(T--)
    {
        scanf("%d%lf" , &n , &p);
        dp[n] = 0;
        for(int i = n - 1 ; i > 0 ;i--)
        {
            double x , y , pp;
            x = (double)i * (n - i);
            y = (double)n  * (n - 1) / 2;
            pp = x / y * p;
            dp[i] = 1.0 * (dp[i + 1] * pp + 1) / pp;
        }
        printf("%.3lf\n", dp[1]);
    }
    return 0;
}