LA4329-乒乓比赛

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

题意：训练指南上的例题。求1到ai-1有多少个数小于ai和ai+1到an中共有多少个数小与ai；

树状数组，从第一个数开始处理，遇到一个数ai，将C[ai]赋值为1，插入树状数组。

#include <iostream>#include <cstring>using namespace std;long long int a[20005],c[20005],d[20005],s[20005];long long int max1;long long int lowbit(long long int x){    return x&-x;}long long int sum(long long int x){   int ret=0;   while(x>0)   {       ret+=c[x];       x-=lowbit(x);   }   return ret;}void add(long long int x,int d1){    while(x<=max1)    {        c[x]+=d1;        x+=lowbit(x);    }}int main(){    int n,t;    max1=-1;    cin>>t;    while(t--)    {      cin>>n;    for(int i=1;i<=n;i++)    {        cin>>a[i];        max1=(max1>a[i])?max1:a[i];    }    memset(c,0,sizeof(c));    for(int i=1;i<=n;i++)    {        s[i]=sum(a[i]);        add(a[i],1);    }    memset(c,0,sizeof(c));    for(int i=n;i>=1;i--)    {        d[i]=sum(a[i]);        add(a[i],1);    }    long long int cnt=0;    for(int i=1;i<=n;i++)    {        cnt+=(s[i]*(n-i-d[i])+(i-1-s[i])*d[i]);    }    cout<<cnt<<endl;    }    return 0;}