POJ2096——概率DP——Collecting Bugs

http://poj.org/problem?id=2096

dp[i][j] 表示 已经侵入了i个子系统，已经拥有了j种类的期望

状态转移方程 dp[i][j] = dp[i][j]*(i*j)/(n*s) + dp[i+1][j]*(n-i)*j/(n*s) + dp[i][j+1]*i*(s-j)/(n*s) + dp[i+1][j+1]*(n-i)*(s-j)/(n*s) + 1

dp[i][j] 可以从已经侵入了i+1个子系统，拥有j种类的情况中删掉一分子系统得到i+1中有n-i中选择情况

dp[i][j] 也可以从dp[i][j]转移，交换

其余同理

/************************************************
* Author        :Powatr
* Created Time  :2015-8-28 16:00:55
* File Name     :C.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 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

double dp[1010][1010];

int main(){
    int n, s;
    while(~scanf("%d%d", &n, &s)){
        memset(dp, 0, sizeof(dp));
        for(int i = n;i >= 0; i--){
            for(int j = s; j >= 0; j--){
                if(i == n && j == s) continue;
                dp[i][j] = dp[i][j+1]*i*(s-j)/n/s + dp[i+1][j]*(n-i)*j/n/s + dp[i+1][j+1]*(n-i)*(s-j)/n/s + 1;
               dp[i][j] /= (1.0 - ((double)i*j/n/s));
            }
        }
        printf("%f
", dp[0][0]);
    }
    return 0;
}