POJ 3071 Football(概率DP)

题目链接

不1Y都对不住看过那么多年的球。dp[i][j]表示i队进入第j轮的概率,此题用0-1<<n表示非常方便。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 #include <cmath>

 5 #include <algorithm>

 6 using namespace std;

 7 double dp[501][501];

 8 double p[501][501];

 9 int main()

10 {

11     int n,i,j,mod,c,k,ans;

12     double maxz;

13     while(scanf("%d",&n)!=EOF)

14     {

15         if(n < 0) break;

16         for(i = 0;i < (1<<n);i ++)

17         {

18             for(j = 0;j < (1<<n);j ++)

19             {

20                 scanf("%lf",&p[i][j]);

21             }

22         }

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

24         {

25             dp[0][i] = 1;

26         }

27         for(i = 1;i <= n;i ++)

28         {

29             for(j = 0;j < (1<<n);j ++)

30             dp[i][j] = 0;

31         }

32         for(i = 1;i <= n;i ++)

33         {

34             for(j = 0;j < (1<<n);j ++)

35             {

36                 mod = j%(1<<i);

37                 if(mod >= (1<<(i-1)))

38                 {

39                     c = (j/(1<<i))*(1<<i);

40                     for(k = 0;k < (1<<(i-1));k ++)

41                     {

42                         dp[i][j] += dp[i-1][j]*dp[i-1][c+k]*p[j][c+k];

43                     }

44                 }

45                 else

46                 {

47                     c = (j/(1<<i))*(1<<i)+(1<<(i-1));

48                     for(k = 0;k < (1<<(i-1));k ++)

49                     {

50                         dp[i][j] += dp[i-1][j]*dp[i-1][c+k]*p[j][c+k];

51                     }

52                 }

53             }

54         }

55         ans = 0;maxz = 0;

56         for(i = 0;i < (1<<n);i ++)

57         {

58             if(maxz < dp[n][i])

59             {

60                 ans = i;

61                 maxz = dp[n][i];

62             }

63         }

64         printf("%d\n",ans+1);

65     }

66     return 0;

67 }