Codeforces Round #320 (Div. 2)(B)

B. Finding Team Member
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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 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 be 2, 1, 4, 3 respectively.

题目看了半天，英语不好就是这样。
贪心每次取最大的数，并将对应的行列变为0

#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int N = 888;int a[N][N];int b[N];int main(void){    int n;    while(~scanf("%d", &n))    {        int i, j, k;        int m = n;        n <<= 1;        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        for(i = 2; i <= n; i++)            for(j = 1; j < i; j++)                scanf("%d", &a[i][j]);        for(k = 1; k <= m; k++)//取m次        {            int max_i = 0, max_j = 0;//得到最大数的下标            for(i = 2; i <= n; i++)            {                for(j = 1; j < i; j++)                {                    if(a[max_i][max_j] < a[i][j])                    {                        max_i = i;                        max_j = j;                    }                }            }            for(i = 1; i <= n; i++)//将对应的行列制为0            {                a[i][max_i] = 0;                a[i][max_j] = 0;                a[max_i][i] = 0;                a[max_j][i] = 0;            }            b[max_i] = max_j;            b[max_j] = max_i;        }        for(i = 1; i <= n-1; i++)            printf("%d ", b[i]);        printf("%d\n", b[n]);    }    return 0;}