HDU 4000 Fruit Ninja 树状数组 + 计数

给你N的一个排列，求满足：a[i] < a[k] < a[j] 并且i < j < k的三元组有多少个。

一步转化：

求出所有满足 a[i] < a[k] < a[j] 并且i < j < k的三元组 与 a[i] < a[k] < a[j] 并且i < k < j的三元组 的个数的总和，设high[i]为在 i 右侧且大于a[i] 的数的个数，上述总和即为：high[i] * (high[i] - 1) / 2。

设low[i]为在 i 左侧且小于a[i] 的数的个数, a[i] < a[k] < a[j] 并且i < k < j的三元组 的个数即为：low[i]*high[i]。

答案即为：∑( high[i] * (high[i] - 1) / 2 - low[i]*high[i] );

 

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <algorithm>

 5 

 6 #define LL long long int

 7 

 8 using namespace std;

 9 

10 const int MAXN = 100100;

11 const LL MOD = 100000007;

12 

13 int N;

14 int C[MAXN];

15 int num[MAXN];

16 int low[MAXN];

17 int high[MAXN];

18 

19 int lowbit( int x )

20 {

21     return x & ( -x );

22 }

23 

24 void Update( int x, int val )

25 {

26     while ( x <= N )

27     {

28         C[x] += val;

29         x += lowbit(x);

30     }

31     return;

32 }

33 

34 int Query( int x )

35 {

36     int res = 0;

37     while ( x > 0 )

38     {

39         res += C[x];

40         x -= lowbit(x);

41     }

42     return res;

43 }

44 

45 int main()

46 {

47     int T, cas = 0;

48     scanf( "%d", &T );

49     while ( T-- )

50     {

51         scanf( "%d", &N );

52         for ( int i = 1; i <= N; ++i )

53             scanf( "%d", &num[i] );

54 

55         memset( C, 0, sizeof(C) );

56 

57         for ( int i = 1; i <= N; ++i )

58         {

59             low[i] = Query( num[i] - 1 );

60             Update( num[i], 1 );

61         }

62 

63         memset( C, 0, sizeof(C) );

64         for ( int i = N; i > 0; --i )

65         {

66             //printf("i=%d %d %d\n", i, Query( N ), Query( num[i] ) );

67             high[i] = Query( N ) - Query( num[i] );

68             Update( num[i], 1 );

69         }

70 

71         LL ans = 0;

72         for ( int i = 1; i <= N; ++i )

73         {

74             //printf( "h=%d l=%d\n", high[i], low[i] );

75             ans += (LL)high[i] * ( high[i] - 1 ) / 2 - (LL)low[i] * high[i];

76             ans = ( ans + MOD ) % MOD;

77         }

78 

79         printf( "Case #%d: %I64d\n", ++cas, ans % MOD );

80     }

81     return 0;

82 }