POJ 2151 Check the difficulty of problems

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3810		Accepted: 1666

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

Sample Output

0.972
题目大意：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。
解题方法：概率dp，dp[i][j][k]表示第i个队在前j道题中解出k道的概率，s[i][k]表示第i队做出的题小于等于k的概率，则

dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k])

s[i][k]=dp[i][M][0]+dp[i][M][1]+``````+dp[i][M][k]

则每个队至少做出一道题概率为P1=(1-s[1][0])*(1-s[2][0])*```(1-s[T][0]);
每个队做出的题数都在1~N-1的概率为P2=(s[1][N-1]-s[1][0])*(s[2][N-1]-s[2][0])*```(s[T][N-1]-s[T][0]);最后的答案就是P1-P2

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

double dp[1010][50][50];
double s[1010][50];
double p[1010][50];

int main()
{
    int M, T, N;
    while(scanf("%d%d%d", &M, &T, &N) != EOF)
    {
        if (M == 0 || N == 0 || T == 0)
        {
            break;
        }
        memset(dp, 0, sizeof(dp));
        memset(s, 0, sizeof(s));
        for (int i = 1; i <= T; i++)
        {
            for (int j = 1; j <= M; j++)
            {
                scanf("%lf", &p[i][j]);
            }
        }
        for (int i = 1; i <= T; i++)
        {
            dp[i][0][0] = 1;//初始化，第i支队伍前0道题做对0道的概率为1
        }
        for (int i = 1; i <= T; i++)
        {
            for (int j = 1; j <= M; j++)
            {
                //第i支队伍前j道题做对0道的概率为dp[i][j - 1][0] * (1.0 - p[i][j])
                dp[i][j][0] = dp[i][j - 1][0] * (1.0 - p[i][j]);
            }
        }
        for (int i = 1; i <= T; i++)
        {
            for (int j = 1; j <= M; j++)
            {
                for (int k = 1; k <= j; k++)
                {
                    //dp[i][j][k]代表第i支队伍前j道题做对k道的概率
                    dp[i][j][k] = dp[i][j - 1][k - 1] * p[i][j] + dp[i][j - 1][k] * (1.0 - p[i][j]);
                }
                //s[i][0]表示第i支队伍做对0道题的概率
                s[i][0] = dp[i][M][0];
                for (int k = 1; k <= M; k++)
                {
                    //求表示第i支队伍做对k道题的概率
                    s[i][k] = dp[i][M][k] + s[i][k - 1];
                }
            }
        }
        double p1 = 1.0;
        double p2 = 1.0;
        for (int i = 1; i <= T; i++)
        {
            p1 *= (1.0 - s[i][0]);//每支队伍至少做对一道题的概率
            p2 *= s[i][N - 1] - s[i][0];//每支队伍最多做对1至N-1道题的概率
        }
        //两个结果相减即为每支队伍至少做对一道题且至少有一支队伍作对的题数在N道以上
        printf("%.3f
", p1 - p2);
    }
    return 0;
}