Codeforces Round #320 (Div. 2) B. 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

261 23 4 5

output

2 1 4 3

input

34870603831 161856845957 794650 97697783847 50566 691206 498447698377 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.

题目大意：

有2n各人组队，人和人之间有对应的友好值，问最优情况时候组队的方式。

题目解法：简单模拟

#include "iostream"#include "string.h"#include "string"#include "queue"#include "cstdio"using namespace std; int s[801][801],n,a[801],vis[801]={0};int main(int argc, char* argv[]){    scanf("%d",&n);    int num=-1,tempx,tempy;    for(int i=2;i<=2*n;i++)    {        for(int j=1;j<=i-1;j++)        {            scanf("%d",&s[i][j]);            if(s[i][j]>num)            {                num=s[i][j];                tempx=i;                tempy=j;            }        }    }    a[tempy]=tempx;    a[tempx]=tempy;    vis[tempx]=1;    vis[tempy]=1;    for(int i=0;i<n-1;i++)    {        num=-1;        for(int i=2;i<=2*n;i++)        {            for(int j=1;j<=i-1;j++)            {                if(!vis[i]&&!vis[j])                {                    if(s[i][j]>num)                    {                        num=s[i][j];                        tempx=i;                        tempy=j;                    }                }            }        }        a[tempy]=tempx;        a[tempx]=tempy;        vis[tempx]=1;        vis[tempy]=1;    }    for(int i=1;i<=2*n;i++)    {        if(i==1)            printf("%d",a[i]);        else            printf(" %d",a[i]);    }    printf("\n");    return 0;}