poj 2151 Check the difficulty of problems （概率DP）

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5006		Accepted: 2203

DescriptionOrganizing 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 youcalculate 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?InputThe 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.OutputFor 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

题意：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。

思路：  dp[i][j][k] 表示第i个对在前j题解出k题的概率，sum[i][j]表示第i个对最多解出j题的概率。

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

sum[i][j]= dp[i][m][0]+dp[i][m][1]+...+ dp[i][m][j];

即  每队至少解出一题的概率 ： ans1 = （1- sum[1][0]）* （1- sum[2][0]）*  ......* （1- sum[m][0]） ;

 每个队解出题但小于n道的概率 ：ans1 = （ sum[m][n-1]- sum[1][0]）* 

  （sum[m][n-1]- sum[2][0]）*  ......* 

（sum[m][n-1]- sum[m][0]） ; 

 所以最后的答案  ans =  ans1-ans2；

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int M=1050;const int N=35;double dp[M],p[M],sum[M];int n,m,t;void input(){for(int i=1;i<=t;i++)for(int j=1;j<=m;j++)scanf("%lf",&p[i][j]);memset(dp,0,sizeof(dp));}void solve(){for(int i=1;i<=t;i++){dp[i][0][0]=1.0;for(int j=1;j<=m;j++)for(int k=0;k<=j;k++){dp[i][j][k]=dp[i][j-1][k]*(1-p[i][j]);if(k!=0) dp[i][j][k]+=dp[i][j-1][k-1]*p[i][j];}sum[i][0]=dp[i][m][0];for(int j=1;j<=m;j++)  sum[i][j]=sum[i][j-1]+dp[i][m][j];}double ans1=1.0,ans2=1.0;for(int i=1;i<=t;i++){ans1*=(1.0-sum[i][0]);ans2*=(sum[i][n-1]-sum[i][0]);}printf("%.3f\n",ans1-ans2);}int main(){while(scanf("%d %d %d",&m,&t,&n)!=EOF){if(m==0 && t==0 && n==0)  break;input();solve();}return 0;}

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