POJ 2096 Collecting Bugs (概率DP)

题意：给定 n 类bug，和 s 个子系统，每天可以找出一个bug，求找出 n 类型的bug，并且 s 个都至少有一个的期望是多少。

析：应该是一个很简单的概率DP，dp[i][j] 表示已经从 j 个子系统中，找出 i 种类型的bug，达到目标所需要天数的期望，

很明显dp[n][s] = 0.0,而dp[0][0] 就是答案，剩下的就比较简单了，

dp[i][j] = (dp[i+1][j]*(n-i)*j + dp[i][j+1]*i*(s-j) + dp[i+1][j+1]*(n-i)*(s-j) + n*s) / (n*s*1.0 - i*j*1.0);

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e3 + 5;

const LL mod = 10000000000007;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

double dp[maxn][maxn];

int main(){

    int s;

    while(scanf("%d %d", &n, &s) == 2){

        dp[n][s] = dp[n+1][s] = dp[n][s+1] = dp[n+1][s+1] = 0.0;

        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+1][j]*(n-i)*j + dp[i][j+1]*i*(s-j) + dp[i+1][j+1]*(n-i)*(s-j) + n*s) / (n*s*1.0 - i*j*1.0);

            }

        printf("%.4f\n", dp[0][0]);

    }

    return 0;

}