CodeForces 579b

　　

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 aredistinct.

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.

Examples

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

#include<string.h>
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct Node
{
    int a,b;
    int wei;
}team[400005];
bool cmp(Node n1,Node n2){
    return n1.wei>n2.wei;
}
int main(){
    int n;
    int ans[805];
    int t=0;
    scanf("%d",&n);
    for(int i=0;i<2*n+10;i++){
        ans[i]=-1;
    }
    for(int i=2;i<=2*n;i++){
        for(int j=1;j<i;j++){
            scanf("%d",&team[t].wei);
            team[t].a=j;
            team[t].b=i;
            t++;
        }
    }
    sort(team,team+t,cmp);
    for(int i=0;i<t;i++){
        if(ans[team[i].a]<0&&ans[team[i].b]<0){
            ans[team[i].a]=team[i].b;
            ans[team[i].b]=team[i].a;

        }
    }
    int flag=0;
    for(int i=1;i<=n*2;i++){
        if(flag) printf(" ");
        printf("%d",ans[i] );
        flag=1;
    }
    printf("
");
    return 0;
}

水题，但是要注意题目给出到是n支队伍，而不是n个人，开始到时候wa了多次，因为没有注意到这一点。