poj 2151 Check the difficulty of problems 概率dp

题意：在acm比赛中，n题，t队。给出每个队做对每题的概率，问每队至少对一题，至少有一队做对至少m题的概率

分析：dp，f[i][j]表示第i个队伍做对第j题的概率。g[i][j][k]表示第i个队伍对于前j题而言做对k道的概率。

g[i][j][k] = g[i][j - 1][k - 1] * (f[i][j]) + g[i][j - 1][k] * (1 - f[i][j]);

有了所有的g，我们就可以求出每个队至少做对1题的概率：ans *= 1 - g[i][n][0];

再减去每个队都只做对1～m-1题的概率(把每个队做对1～m-1题的概率加和，并把各队结果相乘）

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5375   Accepted: 2371 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 2 0.9 0.9 1 0.9 0 0 0 Sample Output 0.972 Source POJ Monthly,鲁小石

#include
#include
#include
#include
using namespace std;
double f[1001][31];
double g[1001][31][31];
int main()
{
    int n,t,m;
    while(~scanf("%d %d %d",&n,&t,&m))
    {
        if(n+t+m==0)
            break;
        for(int i=1; i<=t; i++)
            for(int j=1; j<=n; j++)
                scanf("%lf",&f[i][j]);
        memset(g,0,sizeof(g));
        for(int i=1; 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]);
                }
            }
        }
        double ans=1;
        for(int i=1; i<=t; i++)
            ans*=1-g[i][n][0];
        //ans=1-ans;
        double ans2=1;
        for(int i=1; i<=t; i++)
        {
            double ans1=0;
            for(int j=1; j


不算太难的一道题目，但是错了好几遍原因是poj的%lf要换成%f真心给跪了、。。。。