HDU_5147 Sequence II [逆序数]

最新推荐文章于 2022-01-26 20:08:49 发布

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最新推荐文章于 2022-01-26 20:08:49 发布

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本文介绍了一种使用树状数组求解逆序数的方法来解决SequenceII问题，该问题是寻找符合条件的四元组数量。通过两次遍历序列并利用树状数组计算逆序数，最终得出所有满足条件的四元组总数。

传送门：5147

Sequence II

Problem Description

Long long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence.
Please calculate how many quad (a,b,c,d) satisfy:
1.

1≤a<b<c<d≤n
2.

Aa<Ab
3.

Ac<Ad

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case begins with a line contains an integer n.
The next line follows n integers

A1,A2,…,An .

[Technical Specification]
1 <= T <= 100
1 <= n <= 50000
1 <=

Ai <= n

Output

For each case output one line contains a integer,the number of quad.

Sample Input

Sample Output

思路：树状数组求逆序数

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<string>
#include<utility>
#include<map>
#include<set>
#include<queue>
#include<stack>
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define seed_ 131
#define eps 1e-8
#define mem(a,b) memset(a,b,sizeof a)
#define w(i) tree[i].w
#define ls(i) tree[i].ls
#define rs(i) tree[i].rs
using namespace std;
typedef long long LL;

const int N=50010;
int n;
int a[N],c[N];
LL rev_1[N],rev_2[N];


int lowbit(int x)
{
    return -x&x;
}

void add(int x)
{
    while(x<=n)
    {
        c[x]+=1;
        x+=lowbit(x);
    }
}

LL query(int x)
{
    LL ret=0;
    while(x>0)
    {
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}


int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        mem(c,0);
        mem(rev_1,0);
        mem(rev_2,0);
        scanf("%d",&n);
        LL ans=0;
        int x;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",a+i);
            add(a[i]);
            rev_1[i]=query(n)-query(a[i]);
        }
        mem(c,0);
        for(int i=n;i>=1;i--)
        {
            add(a[i]);
            rev_2[i]=rev_2[i+1]+query(a[i]-1);
        }
//        for(int i=1;i<=n;i++)
//        {
//            printf("%I64d--%I64d\n",rev_1[i],rev_2[i]);
//        }
        for(int i=2;i<=n-2;i++)
        {
            ans+=((LL)i-1-rev_1[i])*((LL)(n-i)*(n-i-1)/2-rev_2[i+1]);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}