bestcoder#23 1002 Sequence II 树状数组+DP

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 652    Accepted Submission(s): 164

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

1 5 1 3 2 4 5

 

Sample Output

4

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 50005+100
const int inf=0x7fffffff;   //无限大
int a[maxn];
ll dp_qmin[maxn];
ll dp2[maxn],sum[maxn];
ll dp_hmax[maxn];
int n;
long long ans;

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

void update(int x,ll val)
{
    while(x <= n)
    {
        sum[x] += val;
        x += lowbit(x);
    }
}

long long query(int x)
{
    long long s=0;
    while(x>0)
    {
        s += sum[x];
        x -= lowbit(x);
    }
    return s;
}

int main()
{
    //freopen("D.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp_qmin[i]=0;
            dp2[i]=0;
            dp_hmax[i]=0;
            sum[i]=0;
        }
        for(int i=1;i<=n;i++){
            ll t=query(a[i]-1);
            dp_qmin[i]=t;
            dp_hmax[i]=(n-i)-(a[i]-1-t);
            //printf("i=%d dpmin=%lld dpmax=%lld\n",i,dp_qmin[i],dp_hmax[i]);
            update(a[i],1);
        }
        ll ans=0;
        ll sum1=0;
        sum1=dp_qmin[1]+dp_qmin[2];
        for(int i=3;i<=n-1;i++)
        {
            ans+=sum1*dp_hmax[i];
            sum1+=dp_qmin[i];
        }
        printf("%I64d\n",ans);
    }
}