UVALive - 4329 Ping Pong (树状数组)

最新推荐文章于 2020-07-30 11:48:58 发布

一只谜谜怪

最新推荐文章于 2020-07-30 11:48:58 发布

阅读量192

点赞数

CC 4.0 BY-SA版权

分类专栏：树状数组

本文链接：https://blog.youkuaiyun.com/qq_39475280/article/details/83218608

树状数组专栏收录该内容

2 篇文章

订阅专栏

本文描述了一道关于乒乓球玩家在一条东西向街道上进行比赛的问题。玩家必须选择一个裁判，且裁判的技能等级不能高于或低于两个参赛者的技能等级。文章提供了一个C++代码解决方案，使用了线段树和差分数组来计算可能的比赛数量。

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2330
题目描述：
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 rst 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 rst integer is N, the number of players. Then
N distinct integers a 1 ; a 2: : : 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

c++代码：


#include <iostream>
#include <cstring>

using namespace std;

const int maxn=2*1e4+5;
const int INF=1e5+5;

typedef long long LL;

int a[maxn];

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

LL c[INF],d[INF];
int n;

int sum(LL x,LL e[]) {
    LL cnt=0;
    while(x>0) {
    	cnt+=e[x];x-=lowbit(x);
    	
	}
	return cnt;
}

void add(LL x,LL y,LL e[]) {
	while(x<=INF) {
		//if(x==1)cout <<e[x] <<endl;
		e[x]+=y;x+=lowbit(x);
	}
}


int main() {
	int T;
	cin >> T;
	while(T--) {
	  memset(d,0,sizeof(d));
	  memset(c,0,sizeof(c));
	  cin >> n;
	   for(int i=1;i<=n;i++) {
		     cin >> a[i];
		     if(i>1)
		     add(a[i],1,d);
	   }
	   LL ans=0;
	   add(a[1],1,c);
	   for(int i=2;i<=n-1;i++) {
	   	     add(a[i],1,c);//记录左边比a[i]小的数的个数
             LL l=sum(a[i]-1,c);
			 LL r=sum(a[i]-1,d);
			 add(a[i],-1,d);
			 //cout << a[i]-1<<" "<<l <<" " << r <<endl;
			 ans+=l*(n-i-r)+r*(i-1-l);	     
	   }
	   cout << ans << endl;
	}	
	return 0;
}