http://poj.org/problem?id=3928

Ping pong
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2087   Accepted: 798

Description

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 first 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 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).

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

Source


题意:每一个人有个能力值,两两即可比赛必须有个裁判,裁判能力值必须在两个人之间位置也必须在两个人之间,问不同的比赛安排有多少种。
简单题解:题里面有句,Now is the problem,脱口而出挖掘机技术哪家强233.。。。
想到枚举裁判即可了,求下每一个人作为裁判的方案有多少种,就是求前面有多少个比它大(小),后面有多少个比它大(小)即可。和求逆序数一样树状数组随便搞搞即可。。。。
代码:
/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=20005;
const int M=100005;
int bit[M];
int f[N];
int a[N];
LL sum(int i)
{
LL s=0;
while(i>0)
{
s+=bit[i];
i-=i&-i;
}
return s;
}
void add(int i,int x)
{
while(i<=M)
{
bit[i]+=x;
i+=i&-i;
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
clr(bit);
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
f[i]=sum(a[i]);
add(a[i],1);
}
clr(bit);
LL ans=0;
for(int i=n-1;i>=0;i--)
{
int x=sum(a[i]);
int y=n-i-1-x;
ans+=(LL)(f[i]*y);
ans+=(LL)((i-f[i])*x);
add(a[i],1);
}
printf("%lld\n",ans);
}
return 0;
}

poj3928 Ping pong 树状数组的更多相关文章

  1. Ping pong(树状数组经典)

    Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  2. UVA 1428 - Ping pong(树状数组)

    UVA 1428 - Ping pong 题目链接 题意:给定一些人,从左到右,每一个人有一个技能值,如今要举办比赛,必须满足位置从左往右3个人.而且技能值从小到大或从大到小,问有几种举办形式 思路: ...

  3. LA 4329 Ping pong 树状数组

    对于我这样一名脑残ACMer选手,这道题看了好久好久大概4天,终于知道怎样把它和“树状数组”联系到一块了. 树状数组是什么意思呢?用十个字归纳它:心里有数组,手中有前缀. 为什么要用树状数组?假设你要 ...

  4. LA4329 Ping pong 树状数组

    题意:一条大街上住着n个乒乓球爱好者,经常组织比赛切磋技术.每个人都有一个能力值a[i].每场比赛需要三个人:两名选手,一名裁判.他们有个奇怪的约定,裁判必须住在两名选手之间,而裁判的能力值也必须在两 ...

  5. UVALive - 4329 Ping pong 树状数组

    这题不是一眼题,值得做. 思路: 假设第个选手作为裁判,定义表示在裁判左边的中的能力值小于他的人数,表示裁判右边的中的能力值小于他的人数,那么可以组织场比赛. 那么现在考虑如何求得和数组.根据的定义知 ...

  6. POJ 3928 Ping pong 树状数组模板题

    開始用瓜神说的方法撸了一发线段树.早上没事闲的看了一下树状数组的方法,于是又写了一发树状数组 树状数组: #include <cstdio> #include <cstring> ...

  7. HDU 2492 Ping pong (树状数组)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2492 Ping pong Problem Description N(3<=N<=2000 ...

  8. LA 4329 - Ping pong 树状数组(Fenwick树)

    先放看题传送门 哭瞎了,交上去一直 Runtime error .以为那里错了. 狂改!!!!! 然后还是一直... 继续狂改!!!!... 一直.... 最后发现数组开小了.......... 果断 ...

  9. hdu 6203 ping ping ping(LCA+树状数组)

    hdu 6203 ping ping ping(LCA+树状数组) 题意:给一棵树,有m条路径,问至少删除多少个点使得这些路径都不连通 \(1 <= n <= 1e4\) \(1 < ...

随机推荐

  1. EF中的事务处理的初步理解

    http://yanwushu.byethost7.com/?p=87 1. EF对事务进行了封装:context的saveChange()是有事务性的. 2. 依赖多个不同的Context的操作(即 ...

  2. c#操作.mpp文件

    原文地址:http://mjm13.iteye.com/blog/532404 所需设置    在工程中增加引用Microsoft Project 11.0 Object Library,该引用在co ...

  3. C++ Primer 学习笔记_79_模板与泛型编程 --模板编译模型

    模板与泛型编程 --模板编译模型 引言: 当编译器看到模板定义的时候,它不马上产生代码.仅仅有在用到模板时,假设调用了函数模板或定义了模板的对象的时候,编译器才产生特定类型的模板实例. 一般而言,当调 ...

  4. Android Studio 入门(转)

    本文适用于从Eclipse转AndroidStudio的开发者 最近打算写一个系列的android初级开发教程,预计40篇以上的文章,结合我实际工作中的经验,写一些工作中经常用到的技术,让初学者可以少 ...

  5. JAVA取整以及四舍五入

    JAVA取整以及四舍五入 import java.math.BigDecimal; //引入这个包 public class Test { public static void main(String ...

  6. Tier和Layer

    在实际开发工作中.我们经常听到"架构设计"和"架构师"这种名词,它并不新奇和神奇,可是却非常少有人对"架构"有全面的了解和认识.更谈不上掌握 ...

  7. hdu 5187 高速幂高速乘法

    http://acm.hdu.edu.cn/showproblem.php?pid=5187 Problem Description As one of the most powerful brush ...

  8. bootstrap之DumpWindowHierarchy

    DumpWindowHierarchy package io.appium.android.bootstrap.handler; import android.os.Environment; impo ...

  9. js检测浏览器中是否安装了flash播放插件

    这两天工作中需要在网页中嵌入flash小游戏,我使用的是swfobject.js version:1.5.其他方面都很好,唯独版本检测这里一直没有搞通,后来实在无奈之下,改用js来检测浏览器的flas ...

  10. CF 439D(251D题)Devu and his Brother

    Devu and his Brother time limit per test 1 second memory limit per test 256 megabytes input standard ...