UVALive - 4329 Ping Pong (树状数组)

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;
}

内容概要:本文系统介绍了算术优化算法(AOA)的基本原理、核心思想及Python实现方法,并通过图像分割的实际案例展示了其应用价值。AOA是一种基于种群的元启发式算法,其核心思想来源于四则运算,利用乘除运算进行全局勘探,加减运算进行局部开发,通过数学优化器加速函数(MOA)和数学优化概率(MOP)动态控制搜索过程,在全局探索与局部开发之间实现平衡。文章详细解析了算法的初始化、勘探与开发阶段的更新策略,并提供了完整的Python代码实现,结合Rastrigin函数进行测试验证。进一步地,以Flask框架搭建前后端分离系统,将AOA应用于图像分割任务,展示了其在实际工程中的可行性与高效性。最后,通过收敛速度、寻优精度等指标评估算法性能,并提出自适应参数调整、模型优化和并行计算等改进策略。; 适合人群:具备一定Python编程基础和优化算法基础知识的高校学生、科研人员及工程技术人员,尤其适合从事人工智能、图像处理、智能优化等领域的从业者;; 使用场景及目标:①理解元启发式算法的设计思想与实现机制;②掌握AOA在函数优化、图像分割等实际问题中的建模与求解方法;③学习如何将优化算法集成到Web系统中实现工程化应用;④为算法性能评估与改进提供实践参考; 阅读建议:建议读者结合代码逐行调试,深入理解算法流程中MOA与MOP的作用机制,尝试在不同测试函数上运行算法以观察性能差异,并可进一步扩展图像分割模块,引入更复杂的预处理或后处理技术以提升分割效果。
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