579B（Finding Team Member）

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.

题解：

按照strength 从大到小选取，一般我是保存下来排序，不过这种方式在数据比较大的时候反而更快一些；

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int person[1000001][2];
int used[1000001];
int ans[810];

int main()
{
int n;
while(scanf("%d", &n) != EOF){
memset(person, 0, sizeof(person));
for(int i = 1; i <= 2*n; ++i)
for(int j = 1; j < i; ++j){
int x;
scanf("%d", &x);
person[x][0] = i;
person[x][1] = j;
}
memset(used, 0, sizeof(used));
used[0] = 1;
for(int i = 1000000; i > 0; --i){
if(!used[person[i][0]] && !used[person[i][1]]){
used[person[i][0]] = used[person[i][1]] = 1;
ans[person[i][0]] = person[i][1];
ans[person[i][1]] = person[i][0];
}
}
for(int i = 1; i <= 2*n; ++i){
if(i > 1) printf(" ");
printf("%d", ans[i]);
}
printf("\n");
}
}