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 aspi.
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 be 0. 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}.

解题说明：此题是一道数学构造题，根据题意矩阵上每个位置上的数是下标为行列的两个数中的最小值。可以考虑先找到每行的最大值，既然该值出现，这个下标为该行号位置上出现的数肯定要大于等于该数，由于数字不重复，一般就是该数，不过由于n是无法选中的，可以随机选中一个行最大值为n-1的位置进行赋值。

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int main()
{
int n,i,j,a[100][100],ans[100],max;
scanf("%d",&n);
for(i=0;i<n;i++)
{
max=0;
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
if(max<a[i][j])
{
max=a[i][j];
}
}
ans[i]=max;
}
for(i=0;i<n;i++)
{
if(ans[i]==n-1)
{
ans[i]=n;
break;
}
}
for(i=0;i<n;i++)
{
printf("%d ",ans[i]);
}
return 0;
}