TCO 2014 Round 1C 概率DP

时间:2014-04-30 19:50:58   收藏:0   阅读:451

TCO round 1C的 250 和500 的题目都太脑残了,不说了。

TCO round 1C 950 一个棋子,每次等概率的向左向右移动,然后走n步之后,期望cover的区域大小?求cover,肯定就是dp[l][r][n], 走了n步之后,左边cover了l,右边cover了r。

一开始DP没有搞清楚,这个要画一下图就更清楚了。 转移方程就是概率的传递方向.

   1:  double dp[505][505][2];  // l,r,n steps unsed;
   2:  class RedPaint
   3:  {
   4:  public:
   5:      double expectedCells(int N)
   6:      {
   7:          memset(dp,0,sizeof(dp));
   8:          // the probability of the configuration.
   9:          dp[0][0][0] = 1.0;
  10:          dp[0][1][1] = 0.5;
  11:          dp[1][0][1] = 0.5;
  12:          for(int n = 2; n<N+1; n++)
  13:          {
  14:              for(int i=0; i<N+1; i++) for(int j=0; j<N+1; j++) dp[i][j][n&1] = 0;
  15:              for(int l= 0; l<N+1; l++)
  16:              {
  17:                  for(int r = 0; r<N+1; r++)
  18:                  {
  19:                      if(l== r && r == 0 ) continue;
  20:                      if( l == 0) dp[l][r][n&1] = (dp[0][r-1][(n-1)&1] + dp[1][r-1][(n-1)&1]) * 0.5;
  21:                      else if(r == 0) dp[l][r][n&1] = (dp[l-1][r][(n-1)&1] + dp[l-1][r+1][(n-1)&1])  * 0.5;
  22:                      else
  23:                          dp[l][r][n&1] = (dp[max(0,l-1)][r+1][(n-1)&1] + dp[l+1][max(r-1,0)][(n-1)&1]) * 0.5;
  24:                  }
  25:              }
  26:          }
  27:          double ret = 0.0f;
  28:          double pro = 0.0f;
  29:          for(int l=0; l<N+1; l++)
  30:          {
  31:              for(int r = 0; r<N+1; r++)
  32:              {
  33:                  pro += dp[l][r][N&1];
  34:                  cout<<l<<" "<<r<<" "<<N<<" "<<dp[l][r][N&1]<<endl;
  35:                  ret += (l+r+1) * dp[l][r][N&1];
  36:              }
  37:          }
  38:          cout<<"All Pro "<<pro<<endl;
  39:          return ret;
  40:      }
  41:  };

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