Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 4522 | Accepted: 1993 |
Description
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
Output
Sample Input
2 2 2 0.9 0.9 1 0.9 0 0 0
Sample Output
0.972
Source
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。假设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,第一遍由于忘记算做出0题的情况,第二遍由于递归中数组开得略大些超内存了。
#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; }
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