Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) 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 p_i. For all pairs of distinct integers i, j between 1 and n, he wrote the number a_i, j = min(p_i, p_j). He writes a_i, i = 0 for all integer i from 1 to n.

Bob gave you all the values of a_i, 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 a_i, j. The j-th number on the i-th line will represent a_i, j. The i-th number on the i-th line will be 0. It's guaranteed that a_i, j = a_j, 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：给你一个矩阵，num[i][j]表示min(a[i],a[j])，求序列a

思路：他们是一个对称矩阵，我们发现，每一行的最大值即为当前序列值，序列可能不唯一

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int num[55][55];
int vis[55];
int ans[55];
int main()
{
	int n,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
		scanf("%d",&num[i][j]);
		int M;
		mem(vis);
		for(i=1;i<=n;i++)
		{
			M=-1;
			for(j=1;j<=n;j++)
			M=max(M,num[i][j]);
			if(vis[M])
			ans[i]=M+1;
			else
			vis[M]=1,ans[i]=M;
		}
		printf("%d",ans[1]);
		for(i=2;i<=n;i++)
		printf(" %d",ans[i]);
		printf("\n");
	}
	return 0;
}