POJ 3928 Ping pong_pingpong! 秋名山-优快云博客

本文针对一个具体的乒乓球竞赛问题，提出了有效的解决策略。该问题要求在给定玩家技能值的情况下，找出所有可能的比赛组合，每场比赛包含两名选手和一名裁判。裁判的技能值必须介于两名选手之间，并且裁判的位置也必须在两选手之间。通过使用树状数组来优化计数过程，文章详细介绍了算法的设计和实现。

Ping pong

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

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.

题目的意思是给你n个乒乓球爱好者的不同能力值，然后组成比赛，选三个人，要求裁判的能力值要在两名选手之间，而且裁判的位置也要在两个选手之间，求最对能组成多少种比赛，每次从1-n枚举当这个人为裁判时，左边有多少比裁判的能力小，右边有多少比裁判小，再用数学乘法处理就知道有多少种了，用树状数组来位数左边和右边的个数，比较巧妙。

#include<stdio.h>
#include<string.h>
using namespace std;
long long c[100005];
long long a[100005];
long long x[100005];
long long y[100005];
int n;
inline int lowbit(int x)
{
    return x&-x;
}
long long sum(int x)
{
    long long ret=0;
    while(x>0)
    {
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}
void add(int x,int d)
{
    while(x<=100005)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}
int main()
{

    int t;
    long long ans;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        scanf("%d",&n);
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            x[i]=sum(a[i]);
            add(a[i],1);
        }
        memset(c,0,sizeof(c));
        for(int i=n;i>=1;i--)
        {
            y[i]=sum(a[i]);
            add(a[i],1);
        }
        for(int i=1;i<=n;i++)
            ans=ans+(x[i]*(n-i-y[i])+(i-x[i]-1)*y[i]);
        printf("%lld\n",ans);
    }
return 0;
}