POJ 3701 概率DP

给定2^n 支足球队进行比赛，n<=7. 队伍两两之间有一个获胜的概率，求每一个队伍赢得最后比赛的概率是多少？

状态其实都是很显然的，一开始觉得这个问题很难啊，不会。dp[i][j] 表示第i支队伍赢得前j轮比赛的概率。(这个题目处理区间的时候比较恶心，小心点即可)。

   1:   

   2:  #include <iostream>

   3:  #include <cstdio>

   4:  #include <cstring>

   5:  #include <map>

   6:  #include <algorithm>

   7:  using namespace std;

   8:   

   9:  double p[135][135];

  10:  double dp[135][10];

  11:   

  12:  pair<int,int> range(int idx, int round)

  13:  {

  14:      int x = (idx>>round) ^1;

  15:      return make_pair(x<<round, (x<<round) + (1<<(round)) - 1);

  16:  }

  17:  int main()

  18:  {

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

  20:      int n;

  21:      while(cin>>n && n !=-1)

  22:      {

  23:          for(int i=0; i<(1<<n); i++)

  24:          {

  25:              for(int j=0; j<(1<<n); j++)

  26:              {

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

  28:              }

  29:          }

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

  31:          for(int i=0; i< (1<<n); i++)

  32:          {

  33:              dp[i][0] = p[i][i^1];

  34:          }

  35:          for(int round=1; round<n; round++) // round

  36:          {

  37:              for(int j=0; j< (1<<n); j++)

  38:              {

  39:                  pair<int,int> r = range(j, round);

  40:                  for(int k = r.first; k<= r.second; k++)

  41:                  {

  42:                      dp[j][round] += p[j][k] *dp[k][round - 1]*dp[j][round-1];

  43:                  }

  44:              }

  45:          }

  46:          int idx  = -1;

  47:          double ret = 0;

  48:          double sum = 0;

  49:          for(int i=0; i<(1<<n); i++)

  50:          {

  51:              sum+= dp[i][n-1];

  52:              if(dp[i][n-1] > ret)

  53:              {

  54:                  idx = i;

  55:                  ret = dp[i][n-1];

  56:              }

  57:          }

  58:          cout<<idx+1<<endl;

  59:      }

  60:  }

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