PAT TOP 1009. Triple Inversions (35)

最新推荐文章于 2023-01-12 20:07:36 发布

原创最新推荐文章于 2023-01-12 20:07:36 发布 · 272 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT TOP

PAT TOP 专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一种高效算法来解决计数三元逆序对的问题，即在一个整数序列中找出所有满足条件i<j<k且Ai>Aj>Ak的三元组数量。文章通过分析给出一种创新的方法，利用数据结构进行优化，最终实现快速计算。

问题描述：

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:

总感觉这一题又像是最大不下降子列的变形题。。。

先对每个输入的数分别求它前面比它大的元素的个数r 和它后面比它小的元素的个数b 两者相乘叠加就AC了；

AC代码（今天学习了平板电视(pb_ds)的相关知识，于是今天的代码变成了(pb_ds)风格的了。。。）

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
int main()
{
//	freopen("data.txt","r",stdin);
	tree<long long,null_type,less<long long>,rb_tree_tag,tree_order_statistics_node_update> tr;
	long long n,x,s=0;
	long long r,b;
	scanf("%lld",&n);
	for(long long i=1;i<=n;i++)
	{
		scanf("%lld",&x);
		tr.insert(x);
		r=tr.size()-tr.order_of_key(x)-1;
		b=r-i+x;
		s=s+r*b;
	}
	printf("%lld",s);
	return 0;
}