# 【BZOJ3943】[Usaco2015 Feb]SuperBull

## Description

Bessie and her friends are playing hoofball in the annual Superbull championship, and Farmer John is in charge of making the tournament as exciting as possible. A total of N (1 <= N <= 2000) teams are playing 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.

## Input

The first line contains the single integer N. The following N lines contain the N team IDs.

## Output

Output the maximum possible number of points that can be scored in the Superbull.

4
3
6
9
10

37

## HINT

FJ先让编号为3和编号为9的队伍进行比赛，然后让编号为9的队伍赢得比赛(淘汰编号为6的队伍)，现在

10的队伍留了下来最后让编号为6和编号为10的队伍比赛，让编号为10的队伍赢得比赛。所有比赛的得分和就是：(

3Xor9)+(6Xor9)+(6Xor10)=10+15+12=37

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,sum;
long long ans;
struct match
{
int from,to,val;
}s[2010*2010];
int p[2010],f[2010];
bool cmp(match a,match b)
{
return a.val>b.val;
}
int find(int x)
{
if(f[x]!=x)
f[x]=find(f[x]);
return f[x];
}
int main()
{
scanf("%d",&n);
int i,j,ra,rb;
for(i=1;i<=n;i++)
{
f[i]=i;
scanf("%d",&p[i]);
for(j=1;j<i;j++)
{
s[m].from=j;
s[m].to=i;
s[m++].val=p[i]^p[j];
}
}
sort(s,s+m,cmp);
for(i=0;i<m;i++)
{
ra=find(s[i].from);
rb=find(s[i].to);
if(ra!=rb)
{
f[ra]=rb;
ans+=(long long)s[i].val;
sum++;
if(sum==n-1)
break;
}
}
printf("%lld",ans);
return 0;
}
posted @ 2017-01-06 16:23  CQzhangyu  阅读(277)  评论(0编辑  收藏  举报