poj Ping pong LA 4329 (树状数组统计数目)

Ping pong

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2302		Accepted: 879

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
讲解：一条大街上住着n个乒乓球爱好者，经常组织比赛，每个人都有一个不同的技能值ai，每场比赛需要三个人，一个裁判，两个队员，有个奇怪的规定，裁判必须住在两名选手中间，并且技能也在两者之间，
求以功能组织多少场比赛；
解：考虑每一个人当裁判的时候，前面大于他的，后面小于他的，前面小于他的，后面大于他的，相乘并相加，然后统计：
AC代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 20010;
const int M = 100010;
int x[M],y[M],ymin[M],ymax[M];
int a[N],lef[N],right[N],leftm[N];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void init( )//统计整体中小于等于i的数共有多少个，表示为ymin[i]
{
    for(int i =1; i<=M; i++)
    {
        ymin[i] = ymin[i-1]+y[i];//n个数中共有多少个小于等于i，以后要减去1；
        ymax[i] = n - ymin[i];//n个数中有多少大于i的；
    }
}
void add(int i,int c)//插入一个数，计算一下后面的
{
    while(i<=M)//第一次提交写了个N，于是wa啦
    {
        x[i] = x[i]+c;
        i=i+lowbit(i);
    }
}
int  solve(int c)
{
    int sum = 0;
    while(c>0)
    {
        sum = sum+x[c];
        c = c-lowbit(c);
    }
    return sum;
}
int main()
{
    int T;
    long long ans;
    scanf("%d",&T);
    while(T--)
    {
        ans = 0;
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        scanf("%d",&n);
        for(int i =1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            y[a[i]] = 1;
        }
        init( );
        for(int i = 1;i<=n;i++)
        {
            add(a[i],1);
            lef[i] = solve(a[i]-1);//求前面小于a[i]的数；
            leftm[i] = i-1-lef[i];//求前面大于a[i] 的数；
            int ma = ymax[a[i]] - leftm[i];//后面大于a[i]的数；
            int mb = ymin[a[i]] -1 - lef[i];//后面小于a[i]的数；
            ans = ans + lef[i]*ma +leftm[i]*mb;//前大后小，前小后大；
        }
        printf("%lld
",ans);
    }
    return 0;
}