TCO 2014 Round 1C 概率DP

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:  };