hdu2429Ping pong

Problem Description

N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment).

Each player has
a unique skill rank. To improve their skill rank, they often compete with each
other. If two players want to compete, they must choose a referee among other
ping pong players and hold the game in the referee's house. For some reason, the
contestants can’t choose a referee whose skill rank is higher or lower than both
of theirs.

The contestants have to walk to the referee’s house, and
because they are lazy, they want to make their total walking distance no more
than the distance between their houses. Of course all players live in different
houses and the position of their houses are all different. If the referee or any
of the two contestants is different, we call two games different. Now is the
problem: how many different games can be held in this ping pong street?

 

Input

The first line of the input contains an integer
T(1<=T<=20), indicating the number of test cases, followed by T lines each
of which describes a test case.

Every test case consists of N + 1
integers. The first integer is N, the number of players. Then N distinct
integers a1, a2 … aN follow, indicating the skill rank of each player, in the
order of west to east. (1 <= ai <= 100000, i = 1 … N).

 

Output

For each test case, output a single line contains an
integer, the total number of different games.

 

Sample Input

1
3 1 2 3

 

Sample Output

1

 

Source

2008
Asia Regional Beijing

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

int an[];
int s[];
const int N = ;
int lb(int x)
{
    return x&(-x);
}
void add(int i)
{
    if(i>N) return;
    s[i]++;
    add(i+lb(i));
}
int get(int i)
{
    if(i == )return ;
    return s[i]+get(i-lb(i));
}
int taa[],tab[],tba[],tbb[];

int main()
{
    int z;
    cin>>z;
    while(z--)
    {
        int n,i,j,k;
        cin>>n;
        memset(s,,sizeof(s));
        memset(taa,,sizeof(taa));
        memset(tab,,sizeof(tab));
        for(i = ;i<=n;i++)
            scanf("%d",&an[i]);
        for(i = ;i<=n;i++)
        {
            taa[i] = get(an[i]-);
            tab[i] = i--get(an[i]);
            add(an[i]);
        }
        memset(s,,sizeof(s));
        memset(tba,,sizeof(tba));
        memset(tbb,,sizeof(tbb));
        for(i = n;i>=;i--)
        {
            tba[i] = get(an[i]-);
            tbb[i] = n-i-get(an[i]);
            add(an[i]);
        }
        long long ans = ;
        for(i = ;i<=n;i++)
        {
            ans += taa[i]*tbb[i]+tab[i]*tba[i];
            ans += (i--taa[i]-tab[i])*(n-i)+(taa[i]+tab[i])*(n-i-tba[i]-tbb[i]);
        }
        cout<<ans<<endl;
    }
    return ;
}