cf #320 B. Finding Team Member（优先队列）

B. Finding Team Member

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

There is a programing contest named SnakeUp, 2n people want to compete for it. In order to attend this contest, people need to form teams of exactly two people. You are given the strength of each possible combination of two people. All the values of the strengths are distinct.

Every contestant hopes that he can find a teammate so that their team’s strength is as high as possible. That is, a contestant will form a team with highest strength possible by choosing a teammate from ones who are willing to be a teammate with him/her. More formally, two people A and B may form a team if each of them is the best possible teammate (among the contestants that remain unpaired) for the other one.

Can you determine who will be each person’s teammate?

Input
There are 2n lines in the input.

The first line contains an integer n (1 ≤ n ≤ 400) — the number of teams to be formed.

The i-th line (i > 1) contains i - 1 numbers ai1, ai2, ... , ai(i - 1). Here aij (1 ≤ aij ≤ 106, all aij are distinct) denotes the strength of a team consisting of person i and person j (people are numbered starting from 1.)

Output
Output a line containing 2n numbers. The i-th number should represent the number of teammate of i-th person.

Sample test(s)

input

2
6
1 2
3 4 5

output

2 1 4 3

input

3
487060
3831 161856
845957 794650 976977
83847 50566 691206 498447
698377 156232 59015 382455 626960

output

6 5 4 3 2 1

Note
In the first sample, contestant 1 and 2 will be teammates and so do contestant 3 and 4, so the teammate of contestant 1, 2, 3, 4will be 2, 1, 4, 3 respectively.

对于这种要频繁求最大值的题，首先想到优先队列。。

要注意，一定要定义cmp函数。。。没有默认。

还有一点。。。优先队列的cmp函数的符号是相反的。。。就是如果要大的在前面。。。那么要定义成 a<b。。。。

/*************************************************************************
> File Name: code/cf/#320/B.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年11月10日 星期二 15时47分29秒
************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int N = 8E2+7;
int n;
int ans
;
struct node
{
int x,y;
int val;
bool operator<(const node &a)const{
return val<a.val;
}
}e;
int main()
{
#ifndef  ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif

ms(ans,-1);
scanf("%d",&n);
priority_queue<node>q;
for ( int i = 2 ; i <= 2*n ; i++)
for ( int j = 1 ; j < i ; j++)
{
int tmp;
scanf("%d",&tmp);
e.x = i;
e.y = j;
e.val = tmp;
q.push(e);
}

while (!q.empty())
{
node pre = q.top();q.pop();
//    cout<<"val:"<<pre.val<<endl;

if (ans[pre.x]==-1&&ans[pre.y]==-1)
{
ans[pre.x] = pre.y;
ans[pre.y] = pre.x;
}
}
for ( int i = 1 ; i <= 2*n ; i++)
if (i!=2*n) printf("%d ",ans[i]);
else printf("%d\n",ans[i]);

#ifndef ONLINE_JUDGE
#endif
fclose(stdin);
return 0;
}

View Code