题目描述

Bessie and her friends are playing hoofball in the annual Superbull championship, and Farmer John isin charge of making the tournament as exciting as possible. A total of N (1 <= N <= 2000) teams areplaying in the Superbull. Each team is assigned a distinct integer team ID in the range 1...2^30-1 to distinguish it from the other teams. The Superbull is an elimination tournament -- after every game, Farmer John chooses which team to eliminate from the Superbull, and the eliminated team can no longer play in any more games. The Superbull ends when only one team remains.Farmer John notices a very unusual property about the scores in matches! In any game, the combined score of the two teams always ends up being the bitwise exclusive OR (XOR) of the two team IDs. For example, if teams 12 and 20 were to play, then 24 points would be scored in that game, since 01100 XOR 10100 = 11000.Farmer John believes that the more points are scored in a game, the more exciting the game is. Because of this, he wants to choose a series of games to be played such that the total number of points scored in the Superbull is maximized. Please help Farmer John organize the matches.贝西和她的朋友们在参加一年一度的“犇”(足)球锦标赛。FJ的任务是让这场锦标赛尽可能地好看。一共有N支球队参加这场比赛,每支球队都有一个特有的取值在1-230-1之间的整数编号(即:所有球队编号各不相同)。“犇”锦标赛是一个淘汰赛制的比赛——每场比赛过后,FJ选择一支球队淘汰,淘汰了的球队将不能再参加比赛。锦标赛在只有一支球队留下的时候就结束了。FJ发现了一个神奇的规律:在任意一场比赛中,这场比赛的得分是参加比赛两队的编号的异或(Xor)值。例如:编号为12的队伍和编号为20的队伍之间的比赛的得分是24分,因为 12(01100) Xor 20(10100) = 24(11000)。FJ相信比赛的得分越高,比赛就越好看,因此,他希望安排一个比赛顺序,使得所有比赛的得分和最高。请帮助FJ决定比赛的顺序

输入

The first line contains the single integer N. The following N lines contain the N team IDs.第一行包含一个整数N接下来的N行包含N个整数,第i个整数代表第i支队伍的编号, 1<=N<=2000

输出

Output the maximum possible number of points that can be scored in the Superbull.一行,一个整数,表示锦标赛的所有比赛的得分的最大值

样例输入

4
3
6
9
10

样例输出

37


题解

由于n只有2000,可以建图然后最大生成树。

有趣的是kruskal都能过

#include <cstdio>
#include <algorithm>
using namespace std;
struct data
{
int x , y , z;
}e[4000001];
int num[2001] , cnt , f[2001];
bool cmp(data a , data b)
{
return a.z > b.z;
}
int find(int x)
{
return x == f[x] ? x : f[x] = find(f[x]);
}
int main()
{
int n , i , j , k = 0;
long long ans = 0;
scanf("%d" , &n);
for(i = 0 ; i < n ; i ++ )
scanf("%d" , &num[i]);
for(i = 0 ; i < n ; i ++ )
{
f[i] = i;
for(j = 0 ; j < n ; j ++ )
{
e[cnt].x = i;
e[cnt].y = j;
e[cnt].z = num[i] ^ num[j];
cnt ++ ;
}
}
sort(e , e + cnt , cmp);
for(i = 0 ; i < cnt && k < n - 1 ; i ++ )
{
int tx = find(e[i].x) , ty = find(e[i].y);
if(tx != ty)
{
k ++ ;
ans += (long long)e[i].z;
f[tx] = ty;
}
}
printf("%lld\n" , ans);
return 0;
}

【bzoj3943】[Usaco2015 Feb]SuperBull的更多相关文章

  1. 【BZOJ3943】[Usaco2015 Feb]SuperBull 最小生成树

    [BZOJ3943][Usaco2015 Feb]SuperBull Description Bessie and her friends are playing hoofball in the an ...

  2. 【BZOJ3943】[Usaco2015 Feb]SuperBull 最大生成树

    [BZOJ3943][Usaco2015 Feb]SuperBull Description Bessie and her friends are playing hoofball in the an ...

  3. 【BZOJ3940】【BZOJ3942】[Usaco2015 Feb]Censoring AC自动机/KMP/hash+栈

    [BZOJ3942][Usaco2015 Feb]Censoring Description Farmer John has purchased a subscription to Good Hoov ...

  4. 【BZOJ3939】[Usaco2015 Feb]Cow Hopscotch 动态规划+线段树

    [BZOJ3939][Usaco2015 Feb]Cow Hopscotch Description Just like humans enjoy playing the game of Hopsco ...

  5. 【BZOJ3940】[USACO2015 Feb] Censoring (AC自动机的小应用)

    点此看题面 大致题意: 给你一个文本串和\(N\)个模式串,要你将每一个模式串从文本串中删去.(此题是[BZOJ3942][Usaco2015 Feb]Censoring的升级版) \(AC\)自动机 ...

  6. 【bzoj3940】[Usaco2015 Feb]Censoring

    [题目描述] FJ把杂志上所有的文章摘抄了下来并把它变成了一个长度不超过10^5的字符串S.他有一个包含n个单词的列表,列表里的n个单词 记为t_1...t_N.他希望从S中删除这些单词.  FJ每次 ...

  7. 【bzoj3942】[Usaco2015 Feb]Censoring

    [题目大意] 有一个S串和一个T串,长度均小于1,000,000,设当前串为U串,然后从前往后枚举S串一个字符一个字符往U串里添加,若U串后缀为T,则去掉这个后缀继续流程. [样例输入] whatth ...

  8. 【bzoj3940】[Usaco2015 Feb]Censoring AC自动机

    题目描述 Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they h ...

  9. 【bzoj3939】[Usaco2015 Feb]Cow Hopscotch 动态开点线段树优化dp

    题目描述 Just like humans enjoy playing the game of Hopscotch, Farmer John's cows have invented a varian ...

随机推荐

  1. ASP.NET 实现上一篇文章 下一篇文章

    select top 1 * from job_hrnews where newsid>162  --下一篇 select top 1 * from job_hrnews where newsi ...

  2. Android开发手记(23) Notification

    有时候,我们需要应用程序在状态内显示一些通知信息,这时我们就需要使用Notification来完成这一工作.也许我们会想到以前经常使用的Toast来通知用户.虽然Notification与Toast都 ...

  3. iOS8中添加的extensions总结(一)——今日扩展

    通知栏中的今日扩展 分享扩展 Action扩展 图片编辑扩展 文件管理扩展 第三方键盘扩展 注:此教程来源于http://www.raywenderlich.com的<iOS8 by Tutor ...

  4. Linux 0.11下信号量的实现和应用

    Linux 011下信号量的实现和应用 生产者-消费者问题 实现信号量 信号量的代码实现 关于sem_wait和sem_post sem_wait和sem_post函数的代码实现 信号量的完整代码 实 ...

  5. 打包静默安装参数(nsis,msi,InstallShield,InnoSetup)[转]

    有时我们在安装程序的时候,希望是静默安装的,不显示下一步下一步,这编访问来教大家如何来操作,现在常用的制作安装程序的软件有,  Microsoft Windows Installer  , Windo ...

  6. Zend Cache的学习和实例

    前一段时间,公司让我组织一下关于Zend Cache的培训. 培训的具体内容有: 前端core缓存 前端Output缓存 前端Function缓存 前端Class缓存 前端File缓存 前端Page缓 ...

  7. mobox:推进企业文档管理走向信息化之路

    随着“大数据”时代的到来,越来越多的人们对数据库管理信息抱有认可态度,这是近年来信息化发展的必然结果.企业作为推进社会经济发展的主力军,也必然要紧跟大数据时代潮流,利用计算机技术全面普及企业的信息化管 ...

  8. SCALA常规练习C

    package com.hengheng.scala abstract class Animal { def walk(speed : Int) def breathe() = { println(& ...

  9. static用法一

    #include "stdafx.h" #include "string.h" struct student { int num; ]; char sex; } ...

  10. Children of the Candy Corn (bfs+dfs)

    Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8120   Accepted: 3547 Description The c ...