Codeforces 618B 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：有一列数字从1到n，数字为pi。 对应着一个n*n的矩阵，在矩阵中 a i,j 表示 min(pi,pj)。  a i,i=0。 现在给出这个矩阵，还原数列。 有多个答案，任意输出一个。

题解： 观察可得在矩阵中第i行出现的最多的数就是数列中pi的值。 对于n，与n-1他们所在的行中所有数都只出现了一次，可以把他们任意放在剩下的两行中。

代码如下：

#include<cstdio>
#include<cstring>
int num[55],ans[55];
int main()
{
	int n,i,j,a,t;
	while(scanf("%d",&n)!=EOF)
	{
		int cnt=0;
		for(i=1;i<=n;++i)
		{
			memset(num,0,sizeof(num));
			for(j=1;j<=n;++j)
			{
				scanf("%d",&a);
				num[a]++;
			}
			t=n+1;
			num[t]=1;
			for(j=1;j<=n;++j)
			{
				if(num[j]>num[t])
					t=j;
			}
			if(t==n+1&&!cnt)
			{
				ans[i]=n;
				cnt=1;
			}
			else if(t==n+1&&cnt)
					ans[i]=n-1;
			else
				ans[i]=t;
		}
		for(i=1;i<=n;++i)
		{
			if(i==n)
				printf("%d\n",ans[i]);
			else
				printf("%d ",ans[i]);
		}
	}
	return 0;
}