UVALive - 4329 Ping pong

最新推荐文章于 2019-04-04 19:48:41 发布

原创最新推荐文章于 2019-04-04 19:48:41 发布 · 377 阅读

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本文探讨了一道关于PingPong竞赛的问题，通过分析N个玩家的技能等级和位置分布，利用树状数组来求解可能的比赛组合数。该问题是算法竞赛中的一道经典题目。

Ping pong

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

N(3 $\le$ N $\le$ 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 $\le$ T $\le$ 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...a_N follow, indicating the skill rank of each player, in the order of west to east ( 1 $\le$ a_i $\le$ 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

Source

Regionals 2008 >> Asia - Beijing

分析：

树状数组题，好好体会这种思想。

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=20000+10;
const int maxnum=100000+10;
int n;
int a[maxn],b[maxn],temp[maxn];
int c[maxnum];

int lowbit(int x)
{
return x&(-x);
}

int sum(int x)//
{
int ret=0;
while(x>0)
{
ret+=c[x];
x-=lowbit(x);
}
return ret;
}

void add(int x,int d)
{
while(x<=maxnum-1)//x<=n就wa了,因为是对1~maxnum进行枚举，不是对n进行枚举（n<=20000）
{
c[x]+=d;
x+=lowbit(x);
}
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&temp[i]);
memset(c,0,sizeof(c));
for(int i=0;i<n;i++)
{
add(temp[i],1);
a[i]=sum(temp[i])-1;//设x[c[i]]=1,所以减去一
}
memset(c,0,sizeof(c));
for(int i=n-1;i>=0;i--)//从后向前修改，注意体会
{
add(temp[i],1);
b[i]=sum(temp[i])-1;
}
long long ans=0;
for(int i=1;i<n-1;i++)
ans+=a[i]*(n-(i+1)-b[i])+(i+1-a[i]-1)*b[i];
printf("%lld\n",ans);
}
return 0;
}