Wunder Fund Round 2016 B. Guess the Permutation

B. Guess the Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bob has a permutation of integers from 1 to n. Denote this permutation as p. The i-th element of p will be denoted as pi. For all pairs of distinct integers i, j between 1 and n, he wrote the number ai, j = min(pi, pj). He writes ai, i = 0 for all integer i from 1 to n.

Bob gave you all the values of ai, j that he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given.

Input

The first line of the input will contain a single integer n (2 ≤ n ≤ 50).

The next n lines will contain the values of ai, j. The j-th number on the i-th line will represent ai, j. The i-th number on the i-th line will be0. It's guaranteed that ai, j = aj, i and there is at least one solution consistent with the information given.

Output

Print n space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print any of them.

Sample test(s)

input

2
0 1
1 0

output

2 1

input

5
0 2 2 1 2
2 0 4 1 3
2 4 0 1 3
1 1 1 0 1
2 3 3 1 0

output

2 5 4 1 3

Note

In the first case, the answer can be {1, 2} or {2, 1}.

In the second case, another possible answer is {2, 4, 5, 1, 3}.

题意：由1 - n组成一个有序数组（p[1],p[2]....p[n]），再每个元素逐个与其他元素比较，输出较小那个，形成矩阵，即矩阵a[i][j] = min (p[i],p[j])，自身位置为0，例如：

样例2中：数组为2 5 4 1 3，那么2逐个与其他比较，取较小那个，就是0 2 2 1 2，也就是矩阵第一行。

思路：除了最大的和第二大的数，也就是n和n-1，其他的数就是取每一行矩阵的最大值，也可以说出现次数最多的值，如果每一个数字都不同，那么肯定就是n或者n-1了，而我这里取巧先做个标记，先取n-1，再取n（看note），若不懂看代码思考一下估计就能明白了，这题就是考数字思维与逻辑（说白就是智商，关键还是看懂题目），不难。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define inf 0x3f3f3f3f

int main()
{
    int n;
    cin>>n;
    int flag=0;
    for(int i=0;i<n;i++)
    {
        int maxn=0;
        for(int j=0;j<n;j++)
        {
            int x;
            cin>>x;
            maxn=max(maxn,x);
        }
        if(maxn==n-1)
        {
            maxn+=flag;
            flag=1;
        }
        cout<<maxn<<" ";
    }
    cout<<endl;
    return 0;
}