POJ 2151 概率DP

主要的子问题是每一个队伍有一个做出题目的概率，求做出k个题目的概率。简单的简单的组合数DP。想清楚即可。

   1:  #include <iostream>

   2:  #include <cstdio>

   3:  #include <cstring>

   4:  using namespace std;

   5:   

   6:  double dp[35][35];

   7:  double p[1005][35];

   8:  int main()

   9:  {

  10:  //    freopen("1.txt","r",stdin);

  11:      int M,T,N;

  12:      while(cin>>M>>T>>N && M!=0)

  13:      {

  14:          memset(p, 0, sizeof(p));

  15:          memset(dp, 0, sizeof(dp));

  16:          for(int i=0; i<T; i++)

  17:          {

  18:              for(int j=0; j<M; j++)

  19:                  cin>>p[i][j];

  20:          }

  21:          double ans = 1.0;

  22:          for(int i=0; i<T; i++)

  23:          {

  24:   

  25:              double ret = 1.0f;

  26:              for(int j=0; j<M; j++)

  27:                  ret *= (1-p[i][j]);

  28:              ret  = 1 - ret;

  29:              ans *= ret;

  30:          }

  31:          double ant = 1.0f;

  32:          for(int i=0; i<T; i++)

  33:          {

  34:              // first i solved num is j

  35:              memset(dp, 0, sizeof(dp));

  36:              dp[1][0] = 1- p[i][0];

  37:              dp[1][1] = p[i][0];

  38:              for(int j=2; j<=M; j++) dp[j][0] = dp[j-1][0] * (1 - p[i][j-1]);

  39:              for(int j=2; j<= M; j++)

  40:              {

  41:                  for(int k = 1; k<=j; k++)

  42:                  {

  43:                      dp[j][k] = dp[j-1][k-1]*(p[i][j-1])+ dp[j-1][k]*(1 - p[i][j-1]);

  44:                  }

  45:              }

  46:              double sum = 0.0f;

  47:              for(int i=1; i<N; i++)

  48:                  sum += dp[M][i];

  49:              ant *= sum;

  50:          }

  51:          printf("%.3f\n", ans - ant);

  52:      }

  53:      return 0;

  54:  }

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