PAT (Top Level) Practise 1009 Triple Inversions (35)

1009. Triple Inversions (35)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CAO, Peng

Given a list of N integers A₁, A₂, A₃,...A_N, there's a famous problem to count the number of inversions in it. An inversion is defined as a pair of indices i < j such that A_i > A_j.

Now we have a new challenging problem. You are supposed to count the number of triple inversions in it. As you may guess, a triple inversion is defined as a triple of indices i < j < k such that A_i > A_j > A_k. For example, in the list {5, 1, 4, 3, 2} there are 4 triple inversions, namely (5,4,3), (5,4,2), (5,3,2) and (4,3,2). To simplify the problem, the list A is given as a permutation of integers from 1 to N.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N in [3, 10⁵]. The second line contains a permutation of integers from 1 to N and each of the integer is separated by a single space.

Output Specification:

For each case, print in a line the number of triple inversions in the list.

Sample Input:

22
1 2 3 4 5 16 6 7 8 9 10 19 11 12 14 15 17 18 21 22 20 13

Sample Output:

8

求三个数的逆序对数，只要求出一个数前后的逆序数相乘即可，直接树状数组都不用离散。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<stack>
#include<algorithm>
#include<bitset>
#include<functional>
using namespace std;
typedef unsigned long long ull;
typedef long long LL;
const int maxn = 1e5 + 10;
const int low(int x){ return x&-x; }
int n, a[maxn], f[maxn], L[maxn], R[maxn];

void add(int x)
{
  for (int i = x; i <= n; i += low(i)) f[i]++;
}

int get(int x)
{
  int ans = 0;
  for (int i = x; i; i -= low(i)) ans += f[i];
  return ans;
}

int main()
{
  while (scanf("%d", &n) != EOF)
  {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) f[i] = 0;
    for (int i = 1; i <= n; i++) L[i] = i - 1 - get(a[i]), add(a[i]);
    for (int i = 1; i <= n; i++) f[i] = 0;
    for (int i = n; i; i--) R[i] = get(a[i]), add(a[i]);
    LL ans = 0;
    for (int i = 1; i <= n; i++) ans += (LL)L[i] * R[i];
    printf("%lld\n", ans);
  }
  return 0;
}