POJ 2151 Check the difficulty of problems （动态规划-概率DP）

Check the difficulty of problems

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4522		Accepted: 1993

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

题目大意：

有 M 道题目 T 支队伍，N表示 最好 的队 至少要做出N题 ，紧接下来T行M列，表示某队做出某题 的概率为p ，问你每支队至少做出1题，最好的队至少做出N题的概率是多少？

解题思路：

一题动态规划的题， 既然最好的队至少做出N题，那么用二维记录，DP [t][f] 记录还剩 t 支队及是否出现超过N题的事件的概率。如果当前这支队伍做出超过N题，那么f置为1，否则还是f。弹了两遍，第一遍因为忘记算做出0题的情况，第二遍因为递归中数组开得略大些超内存了。

解题代码：

#include <iostream>
#include <cstdio>
using namespace std;

const int maxt=1100;
const int maxn=32;
double dp[maxt][2];
double p[maxt][maxn];

int T,M,N;

double DP(int t,int f){
    if(t<=0) return f;
    if(dp[t][f]>-1.0) return dp[t][f];

    double a[maxn][maxn],ans=0;
    a[0][0]=1;
    for(int i=1;i<=M;i++){
        for(int j=0;j<=i;j++){
            a[i][j]=0;
            if(j-1>=0) a[i][j]+=a[i-1][j-1]*p[t][i];
            if(i-1>=j) a[i][j]+=a[i-1][j]*(1-p[t][i]);
        }
    }
    for(int i=1;i<N;i++) ans+=DP(t-1,f)*a[M][i];
    for(int i=N;i<=M;i++) ans+=DP(t-1,1)*a[M][i];

    return dp[t][f]=ans;
}

void input(){
    for(int i=1;i<=T;i++){
        dp[i][0]=dp[i][1]=-2.0;
        for(int j=1;j<=M;j++){
            scanf("%lf",&p[i][j]);
        }
    }
}

void solve(){
    printf("%.3f\n",DP(T,0));
}

int main(){
    while(scanf("%d%d%d",&M,&T,&N)!=EOF && (T||M||N) ){
        input();
        solve();
    }
    return 0;
}