POJ 3928-Ping pong（樹狀陣列+加/乘法原理）

kewlgrl發表於2016-08-08
陣列

Ping pong
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2982   Accepted: 1103

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

題目意思：

一條街上有n個乒乓球愛好者，每個人都有一個不同的技能值 a 。每場比賽需要選3個人：1個裁判，2個選手，有一個奇怪的規定：裁判必須住在選手中間，且裁判的技能值也要在選手中間。求能有多少種比賽。

解題思路：（來源已未可考）

列舉第 i 個人當裁判，設 a0 到 ai-1 中，有 bi 個數比 ai 小，ai+1 到 an-1中有 di 個數比 ai 大，那麼比賽種數為 b[i]*(n-i-1-d[i])+(i-b[i])*d[i].

要求 bi 和 di ，可以採用樹狀陣列。ai 的最大值為100000，那麼BIT的範圍是1-100000，A[1,100000]每個數 A[i] 代表一個人，一開始BIT中的每個位置上的數都是0，表示這個位置上沒有人，C[] 表示BIT中的輔助陣列，初始化為0。每當插入一個 ai ，相當於把值為 ai 的人放在BIT中的相應位置上，即對應位置上的值要加1，從初始的0變為1，此時需要修改相應的C[] 中的值。

由於我要求的是比 ai 小的數的個數，那麼這些比 ai 小的數在BIT中的位置必然是在 ai 之前，也就是這些位置上的值為1，那麼 A[ai] 的字首和就等於這些數的個數。

當要求 ai+1 到 an-1 中比 ai 小的數時，把陣列 a[] 反過來插入BIT中即可。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <cstdlib>
#define maxn 100010
#define ll long long
using namespace std;
int a[maxn],c[maxn];
int mins[maxn],maxs[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int d)
{
    while(x<=maxn)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}

int sum(int i)
{
    int res=0;
    while(i>0)
    {
        res+=c[i];
        i-=lowbit(i);
    }
    return res;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        memset(c,0,sizeof(c));
        int n;
        cin>>n;
        for(int i=1; i<=n; ++i)
            cin>>a[i];
        for(int i=1; i<=n; ++i)
        {
            mins[i]=sum(a[i]);//計算比a[i]小的數的個數
            update(a[i],1);//放進一個a[i]
        }
        memset(c,0,sizeof(c));
        for(int i=n; i>0; --i)
        {
            maxs[i]=sum(a[i]);
            update(a[i],1);
        }
        ll ans=0;
        for(int i=1; i<=n; i++)
            ans+=(mins[i]*(n-i-maxs[i])+(i-1-mins[i])*maxs[i]);
        cout<<ans<<endl;
    }
    return 0;
}
/**
1
3
1 2 3
**/



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