Codeforces Round #540 (Div. 3) C.Palindromic Matrix

http://codeforces.com/contest/1118/problem/C
题意就是给$n^2$个数，填成一个上下左右都对称的矩阵。
submission

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#define LL long long
#define N 1010
using namespace std;
int n;
int num[N],mtx[25][25],odd,odd_num;
inline void init(){
    int t;
    scanf("%d",&n);
    memset(num,0,sizeof num);
    for (int i = 1;i <= n*n; i++)
    {
        scanf("%d",&t);
        num[t]++;
    }
}
int check0(){
    for (int i = 1;i <= 1000; i++)
        if (num[i]%4) return 0;
    return 1;
}
int check1(){
    int cnt = 0;
    odd = 0;
    for (int i = 1; i <= 1000; i++)
    {
        if (num[i]>=4) cnt+=(num[i]/4)*4;
        if (num[i]&1) 
        {
            odd++;
            odd_num = i;
        }
        if (odd>1) return 0;
    }
    if (cnt < (n-1)*(n-1)) return 0;
    return 1;
}
inline void fill0(){
    for (int i = 1; i <= n/2; ++i)
        for (int j = 1; j <= n/2; ++j)
    {
        for (int k = 1; k <= 1000; k++)
            if (num[k])
        {
            mtx[i][j] = k;
            mtx[n-i+1][j] = k;
            mtx[i][n-j+1] = k;
            mtx[n-i+1][n-j+1] = k;
            num[k]-=4;
            break;
        }
    }
}
inline void fill1(){
    if (odd) 
    {
        mtx[n/2+1][n/2+1] = odd_num;
        num[odd_num]--;
    }
    for (int i = 1; i <= n/2; ++i)
        for (int j = 1; j <= n/2; ++j)
    {
        for (int k = 1; k <= 1000; k++)
            if (num[k] && num[k]>=4)
        {
            mtx[i][j] = k;
            mtx[n-i+1][j] = k;
            mtx[i][n-j+1] = k;
            mtx[n-i+1][n-j+1] = k;
            num[k]-=4;
            break;
        }
    }    
    for(int i = 1; i <= n/2; i++)
    {
        for (int k = 1; k <= 1000; k++)
            if (num[k])
        {
            mtx[i][n/2+1] = k;
            mtx[n-i+1][n/2+1] = k;
            num[k]-=2;
            break;
        }
    }
    for(int j = 1; j <= n/2; j++)
    {
        for (int k = 1; k <= 1000; k++)
            if (num[k])
        {
            mtx[n/2+1][j] = k;
            mtx[n/2+1][n-j+1] = k;
            num[k]-=2;
            break;
        }
    }
}
inline void print(){
    printf("YES\n");
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<n;j++) printf("%d ",mtx[i][j]);
        printf("%d\n", mtx[i][n]);
    }
}
int main(){
    init();
    if (n&1)
    {
        if (!check1()) printf("NO\n");
        else 
        {
            fill1();//n为奇数
            print();
        }
    }   
    else 
    {
        if (!check0()) printf("NO\n");
        else 
        {
            fill0();//n为偶数
            print();
        }
    }
    return 0;
}