POJ 2151 Check the difficulty of problems （動態規劃-機率DP）

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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 20.9 0.91 0.90 0 0

Sample Output

0.972

Source

POJ Monthly,魯小石

題目大意：

有 M 道題目 T 支隊伍，N表示 最好 的隊 至少要做出N題 ，緊接下來T行M列，表示某隊做出某題 的機率為p ，問你每支隊至少做出1題，最好的隊至少做出N題的機率是多少？

解題思路：

一題動態規劃的題， 既然最好的隊至少做出N題，那麼用二維記錄，DP [t][f]  記錄還剩 t 支隊及是否出現超過N題的事件的機率。如果當前這支隊伍做出超過N題，那麼f置為1，否則還是f。彈了兩遍，第一遍因為忘記算做出0題的情況，第二遍因為遞迴中數組開得略大些超記憶體了。

解題代碼：

 

#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;}