[Wunder Fund Round 2016(Div 1 + Div 2 combined) B] [폭력 욕심] Guess the Permutation 전체 배열 a[i] [j]=min(p[i],p

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.
Examples
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}.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
int a[55][55];
int val[55];
int main()
{
	while (~scanf("%d", &n)) 
	{
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= n;++j)	{
				scanf("%d", &a[i][j]);
			}
		}
		MS(val, 0);
		for (int v = 1; v <= n; ++v)
		{
			for (int i = 1; i <= n; ++i)if(val[i]==0)
			{
				bool flag = 1;
				for (int j = 1; j <= n; ++j)
				{
					if (a[i][j] > v)flag = 0;
				}
				if (flag)
				{
					val[i] = v;
					break;
				}
			}
		}
		for (int i = 1; i <= n; ++i)printf("%d ", val[i]);
		puts("");
	}
	return 0;
}
/*
【  】
    n（2<=n<=50），         n    p[]。
  a[i][j]=min(p[i],p[j])，i!=j
  i==j，  a[i][j]=0
   a[][]，   p[]

【  】
  

【  】
            ，    1,2,3,...       。

【     &&  】
O（n^3）

*/