Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] B. Finding Team Member 排序

                                                                  B. Finding Team Member

                                                                              time limit per test

2 seconds

                                                                              memory limit per test

256 megabytes

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 a_i1, a_i2, ... , a_i(i - 1). Here a_ij (1 ≤ a_ij ≤ 10⁶, all a_ij 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

题解：没脑子的找就是了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
#define maxn 800+6
vector< pair<int ,pair<int ,int > > > mp;
set<int >s;
int ans[maxn];
int main()
{

   int n,x;
   cin>>n;
    for(int i=;i<=*n;i++)
    {
        for(int j=;j<i;j++)
        {
          cin>>x;
            mp.push_back(make_pair(x,make_pair(i,j)));
        }
    }
    sort(mp.begin(),mp.end());
    int k=mp.size();
    for(int i=k-;i>=;i--)
    {
        if(s.count(mp[i].second.first)||s.count(mp[i].second.second))continue;
        s.insert(mp[i].second.first);
        s.insert(mp[i].second.second);
        ans[mp[i].second.first]=mp[i].second.second;
        ans[mp[i].second.second]=mp[i].second.first;
    }
    for(int i=;i<=*n;i++)
        cout<<ans[i]<<" ";
    return ;
}

代码