Ping pong
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3560 Accepted Submission(s): 1312
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?
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).
1 3 1 2 3
1
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn=100010;
int sum[maxn<<2],dpl[maxn],dpr[maxn],val[maxn];
void update(int L,int R,int p,int k)
{
int ls,rs,mid;
if(L==R)
{
sum[k]=1;
return ;
}
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
if(p<=mid)
update(L,mid,p,ls);
else
update(mid+1,R,p,rs);
sum[k]=sum[ls]+sum[rs];
}
int qu(int L,int R,int l,int r,int k)
{
int ls,rs,mid;
if(l==L&&r==R)
return sum[k];
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
if(l>mid)
return qu(mid+1,R,l,r,rs);
else if(r<=mid)
return qu(L,mid,l,r,ls);
else
return qu(L,mid,l,mid,ls)+qu(mid+1,R,mid+1,r,rs);
}
int main()
{
int t,n,i,lim;
scanf("%d",&t);
while(t--)
{
lim=0;memset(sum,0,sizeof sum);
scanf("%d",&n);
memset(sum,0,sizeof sum);
for(i=0;i<n;i++)
scanf("%d",&val[i]),lim=max(lim,val[i]);
for(i=0;i<n;i++)
{
dpl[i]=qu(1,lim,1,val[i],1);
update(1,lim,val[i],1);
}
memset(sum,0,sizeof sum);
for(i=n-1;i>=0;i--)
{
dpr[i]=qu(1,lim,val[i],lim,1);
update(1,lim,val[i],1);
}
long long ans=0;
for(i=0;i<n;i++) ans+=dpl[i]*dpr[i]+(i-dpl[i])*(n-i-1-dpr[i]);
cout<<ans<<endl;
}
return 0;
}