poj 2151 Check the difficulty of problems（dp）

题目链接

题目描述：
Check the difficulty of problems
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 6668 Accepted: 2899
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

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

题解：
设dp[i][j][k]表示第i个队在前j道题中解出k道的概率
则：
dp[i][j][k]=dp[i][j-1][k-1]p[j][k]+dp[i][j-1][k](1-p[j][k]);
先初始化算出dp[i][0][0]和dp[i][j][0];
设s[i][k]表示第i队做出的题小于等于k的概率
则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<iostream>#include<cstdio>#include<algorithm>using namespace std;const int maxn=30+5;const int Maxn=1000+5;double d[Maxn][maxn][maxn],s[Maxn][maxn],p[Maxn][maxn];int main(){    //freopen("in.txt","r",stdin);    int t,n,m;    while(~scanf("%d%d%d",&t,&n,&m))    {        if(t==0&&n==0&&m==0)        break;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=t;j++)            scanf("%lf",&p[i][j]);        }        /*for(int i=1;i<=n;i++)        {            d[i][0][0]=1;            for(int j=1;j<=t;j++)            {                d[i][j][0]=d[i][j-1][0]*(1-p[i][j]);            }        }*/        for(int i=1;i<=n;i++)        {            d[i][0][0]=1;            for(int j=1;j<=t;j++)            d[i][0][j]=0;            d[i][1][0]=1-p[i][1];            for(int j=1;j<=t;j++)            d[i][j][0]=d[i][j-1][0]*(1-p[i][j]);            for(int j=1;j<=t;j++)            {                for(int k=1;k<=j;k++)                {                    d[i][j][k]=d[i][j-1][k-1]*p[i][j];                    if(j-1>=k)                    d[i][j][k]+=d[i][j-1][k]*(1-p[i][j]);                }            }        }        for(int i=1;i<=n;i++)        {            //s[i][0]=d[i][t][0];            /*for(int j=1;j<=t;j++)            s[i][j]=s[i][j-1]+d[i][t][j];*/            for(int j=0;j<=t;j++)            {                s[i][j]=0;                for(int k=0;k<=j;k++)                s[i][j]+=d[i][t][k];            }        }        double p1=1,p2=1;        for(int i=1;i<=n;i++)        {            p1*=(1-s[i][0]);        }        for(int i=1;i<=n;i++)        {            p2*=(s[i][m-1]-s[i][0]);        }        printf("%.3lf\n",p1-p2);    }}