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;
}