Tag Archives: 概率

POJ 2151-Check the difficulty of problems

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POJ 2151-Check the difficulty of problems Check the difficulty of problems Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5066 Accepted: 2238 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

这题纯粹的数学概率题，不太好做。题目大意是：有m个问题，t个队，要求冠军至少解决n个问题，所有队至少解决1个问题，问这样的概率是多少。对于题目中给的样例，我们可以手算得出，分3种情况：a.第一个队是冠军，第二个队只解出一题0.9*0.9*1*0.1=0.081；b.第二个队是冠军，第一个队只解出一题(0.9*0.1+0.1*0.9)*1*0.9=0.162；c.两个队都是冠军，也就是并列冠军0.9*0.9*1*0.9=0.729；把这三种情况加起来就是0.972。 但是如果m,t的数字很大，问题就没那么简单了，解题思路比较难想到，可以参考以下两篇博客： 比较难理解的是概率的转化：
要求：每队至少解出一题且冠军队至少解出N道题的概率。
由于冠军队可以不止一队，即允许存在并列冠军。 则原来的所求的概率可以转化为：每队均至少做一题的概率P1 减去 每队做题数均在1到N-1之间的概率P2。 如果理解了上面那句话，使用概率DP就很容易写出代码： [cpp] #include<iostream> #include<cstdio> using namespace std; int m,t,n; double p[1001][31];//p[i][j]第i个队解出第j题的概率 double dp[1001][31][31];//dp[i][j][k]第i个队在前j个题中解出k个题的概率dp[i][j][k]=dp[i][j-1][k-1]*p[j][k]+dp[i][j-1][k]*(1-p[j][k])分第j个题能和不能解出来 double s[1001][31];//s[i][k]第i个队最多解出k个题的概率s[i][k]=dp[i][M][0]+dp[i][M][1]+…+dp[i][M][k] void solve() { for(int i=1;i<=t;i++) { dp[i][0][0]=1.0;//初始化，第i支队伍前0道题做对0道的概率为1 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]; } } s[i][0]=dp[i][m][0]; for(int j=1;j<=m;j++) s[i][j]=s[i][j-1]+dp[i][m][j]; } double p1=1.0,p2=1.0; for(int i=1;i<=t;i++) { p1*=(1-s[i][0]); p2*=(s[i][n-1]-s[i][0]); } printf("%.3f\n",p1-p2); } int main() { while(cin>>m>>t>>n&&m&&t&&n) { for(int i=1;i<=t;i++) for(int j=1;j<=m;j++) cin>>p[i][j]; solve(); } return 0; } [/cpp] 本代码提交AC，用时94MS，内存8228K。]]>