POJ 3928 & HDU 2492 Ping pong（树状数组求逆序数）

题目链接：

PKU:http://poj.org/problem?id=3928

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2492

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

Beijing 2008

PS:

分别求每一个数的左右两边比它大的个数和小的个数！最后再交叉相乘即可！

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=100017;
int n;
int a[maxn], c[maxn];
int leftMax[maxn], leftMin[maxn];
int rightMax[maxn], rightMin[maxn];
typedef __int64 LL;

int Lowbit(int x) //2^k
{
    return x&(-x);
}

void update(int i, int x)//i点增量为x
{
    while(i <= maxn)//注意此处
    {
        c[i] += x;
        i += Lowbit(i);
    }
}
int sum(int x)//区间求和 [1,x]
{
    int sum=0;
    while(x>0)
    {
        sum+=c[x];
        x-=Lowbit(x);
    }
    return sum;
}

int main()
{
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        memset(c,0,sizeof(c));
        for(int i = 1; i <= n; i++)
        {
            leftMin[i] = sum(a[i]);//计算左边小的个数
            leftMax[i] = i-leftMin[i]-1;//计算左边大的个数
            update(a[i],1);
        }
        memset(c,0,sizeof(c));//再次清零
        for(int i = n,j = 1; i >= 1; i--,j++)
        {
            rightMin[i] = sum(a[i]);
            rightMax[i] = j-rightMin[i]-1;
            update(a[i],1);
        }
        LL ans = 0;
        for(int i = 1; i <= n; i++)
        {
            ans+=leftMax[i]*rightMin[i] + leftMin[i]*rightMax[i];//交叉相乘取和
        }
        printf("%I64d\n",ans);
    }
    return 0;
}