CF div2 320 B

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.

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, 4 will be2, 1, 4, 3 respectively.

题目里说保证两个人的组队的数值不同，用一个二位数组a[i][0] ，a[i][1]分别记录i这个数出现的行数跟列数，然后从大到小跑一遍就行了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

const int M = 1000005;

int a[M][2];
int ans[1050];
int vis[1024];

int main()
{
int n;
scanf("%d",&n);
n*=2;
int x;
for(int i=2;i<=n;i++){
for(int j=1;j<i;j++){
scanf("%d",&x);
a[x][0]  = i;
a[x][1] = j;
}
}
vis[0] = 1;
for(int i=M;i>=0;i--){
if(!vis[a[i][0]]&&!vis[a[i][1]]){
vis[a[i][0]] = 1;
vis[a[i][1]] = 1;
ans[a[i][0]] = a[i][1];
ans[a[i][1]] = a[i][0];
}
}
for(int i=1;i<=n;i++){
printf("%d ",ans[i]);
}
printf("\n");

return 0;
}

View Code