POJ 2151 Check the difficulty of problems 概率DP

ACM比赛中，共M道题，T个队，pij表示第i队解出第j题的概率 问 每队至少解出一题且冠军队至少解出N道题的概率。分析： 对于需要求得概率比较容易想到： 假设p1为每个队至少解出一题的概率，这个容易算出。 假设p2为每个队至少解出一题但是不超过n-1题的概率 所以最终答案为：p1-p2 现在问题是如何求出p2? 假设dp[i][j]表示第i个队解出的题目<=j的概率 则dp[i][j]=解出1题+解出2题+...解出j题的概率 现在问题转化为如何求解出1，2，3...k题的概率 假设g[i][j][k]表示第i个队在前j题解出k题的概率则:g[i][j][k]=g[i][j-1][k-1]*p[i][j]+g[i][j-1][k]*(1-p[i][j]);所以x[i][M][k]表示的就是第i个队解出k题的概率 假设
Check the difficulty of problems
Time Limit: 2000MSMemory Limit: 65536KTotal Submissions: 6540Accepted: 2845
Description
Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms:1. All of the teams solve at least one problem.2. The champion (One of those teams that solve the most problems) solves at least a certain number of problems.Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem.Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability Pij (1 <= i <= T, 1<= j <= M). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems?
Input
The input consists of several test cases. The first line of each test case contains three integers M (0 < M <= 30), T (1 < T <= 1000) and N (0 < N <= M). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just Pij. A test case of M = T = N = 0 indicates the end of input, and should not be processed.
Output
For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.
Sample Input
2 2 20.9 0.91 0.90 0 0
Sample Output
0.972
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>usingnamespace std;#define maxn 35#define maxt 1005int n, t, m;double f[maxt][maxn];double g[maxt][maxn][maxn];int main(){    //freopen("t.txt", "r", stdin);while (scanf("%d%d%d", &n, &t, &m), n | t | m)    {        double ans =1;        for (int i =0; i < t; i++)            for (int j =1; j <= n; j++)                scanf("%lf", &f[i][j]);        memset(g, 0, sizeof(g));        for (int i =0; i < t; i++)        {            g[i][0][0] =1;            for (int j =1; j <= n; j++)            {                g[i][j][0] = g[i][j -1][0] * (1- f[i][j]);                for (int k =1; k <= j; k++)                    g[i][j][k] = g[i][j -1][k -1] * (f[i][j])                            + g[i][j -1][k] * (1- f[i][j]);            }        }        for (int i =0; i < t; i++)            ans *=1- g[i][n][0];        double temp =1;        for (int i =0; i < t; i++)        {            double sum =0;            for (int j =1; j < m; j++)                sum += g[i][n][j];            temp *= sum;        }        ans -= temp;        printf("%.3f\n", ans);    }    return0;}
 

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