Description
N(3
N20000) 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(1T20) ,
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 ( 1ai100000 , 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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int max(int a,int b) {return a>b?a:b;}
inline int lowbit(int x)
{
return x&(-x);//2^k
}
struct Fen
{
int n;
vector<int> c;
void resize(int n) {this->n=n;c.resize(n);}
void clear() {fill(c.begin(),c.end(),0);}
int sum(int x) //求前 N项的和
{
int r=0;
while(x>0)
{
r+=c[x];
x-=lowbit(x);
}
return r;
}
void add(int x,int d)//求前 N项sum的和
{
while(x<=n)
{
c[x]+=d;x+=lowbit(x);
}
}
};
int n,a[20005],c[20005],d[20005];
Fen f;
int main()
{
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(a,0,sizeof(a));
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
int maxs=0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
maxs=max(maxs,a[i]);
}
f.resize(maxs+1);
f.clear();
for(i=1;i<=n;i++)//前缀和
{
f.add(a[i],1);//求前 N项sum的和
c[i]=f.sum(a[i]-1);//求前 N项的和
}
f.clear();
for(i=n;i>=1;i--)//后缀和
{
f.add(a[i],1);
d[i]=f.sum(a[i]-1);
}
long long ans=0;
for(i=2;i<n;i++)
ans+=(long long)c[i]*(n-i-d[i])+(long long)d[i]*(i-1-c[i]);
printf("%lld\n",ans);
}
return 0;
}
原文地址：http://blog.csdn.net/u012346225/article/details/42128255
