Wunder Fund Round 2016 (Div. 1 + Div. 2 combined)-B. Guess the Permutation（模拟）

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}.

思路：
   这题其实只要将每列出现次数最多的数统计出来就是这个位置的数了，如果每个数都是出现一样多的话，那么就是可以随便填，所以将固定位置的数填好后，其他位置就随便找一个没填过的数放入就行了。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
using namespace std;
const int T=150;
#define inf 0x3f3f3f3fL
#define mod 1000000000
typedef long long ll;
typedef unsigned long long LL;

int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif

int n,m,i,j,k;
bool vis[T];
int v[T][T],cnt[T][T];
while(~scanf("%d",&n))
{
memset(vis,false,sizeof(vis));
memset(cnt,0,sizeof(cnt));
for(i=0;i<n;++i){
for(j=0;j<n;++j){
scanf("%d",&v[i][j]);
cnt[i][v[i][j]]++;
if(cnt[i][51]<cnt[i][v[i][j]]){
cnt[i][51] = cnt[i][v[i][j]];
cnt[i][52] = v[i][j];
}
}
}
for(i=0;i<n;++i){
if(cnt[i][51]>1){
vis[cnt[i][52]]=true;
}
}
for(i=0;i<n;++i){
if(cnt[i][51]>1){
printf("%d ",cnt[i][52]);
}
else {
for(j=1;j<=n;++j){
if(!vis[j]){
printf("%d ",j);
vis[j] = true;
break;
}
}
}
}
printf("\n");
}
return 0;
}