POJ 2096 Collecting Bugs (概率DP)
Posted dwtfukgv
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 2096 Collecting Bugs (概率DP)相关的知识,希望对你有一定的参考价值。
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; }
以上是关于POJ 2096 Collecting Bugs (概率DP)的主要内容,如果未能解决你的问题,请参考以下文章