POJ 3435 Sudoku Checker

POJ 3435 Sudoku Checker

Description

The puzzle game of Sudoku is played on a board of N² × N² cells. The cells are grouped in N × N squares of N × N cells each. Each cell is either empty or contains a number between 1 and N².

The sudoku position is correct when numbers in each row, each column and each square are different. The goal of the game is, starting from some correct position, fill all empty cells so that the final position is still correct.

This game is fairly popular in the Internet, and there are many sites which allow visitors to solve puzzles online. Such sites always have a subroutine to determine a correctness of a given position.

You are to write such a routin.

Input

Input file contains integer N, followed by N⁴ integers — sudoku position. Empty cells are denoted by zeroes.

Constraints

1 ≤ N ≤ 10.

Output

Output file must contain a single string 'CORRECT' or 'INCORRECT'.
　

Sample Input

Sample input 1
2
0 0 0 0
0 0 0 0
0 0 2 0
0 0 0 1
Sample input 2
2
2 1 3 0
3 2 4 0
1 3 2 4
0 0 0 1

Sample Output

Sample output 1
CORRECT
Sample output 2
INCORRECT

臥槽，數獨，判斷行，列和小方格裡的數不重復。但只判斷當前狀態。

#include
#include
#include
#include
#include
using namespace std;
const int maxn=110;
int a[maxn][maxn],n;
int x[maxn][15];
int y[maxn][15];
int s[maxn][15];

int main()
{
    while(~scanf("%d",&n))
    {
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        memset(s,0,sizeof(s));
        bool flag=true;
        for(int i=0;i0&&flag)
                {
                    if(!x[i][a[i][j]])//判斷行
                       x[i][a[i][j]]=1;
                    else
                       flag=false;
                    if(!y[j][a[i][j]])//判斷列
                       y[j][a[i][j]]=1;
                    else
                        flag=false;
                    if(!s[i/n*n+j/n+1][a[i][j]])//判斷小方格
                        s[i/n*n+j/n+1][a[i][j]]=1;
                    else
                        flag=false;
                }
            }
        }
        if(flag)
            printf("CORRECT\n");
        else
            printf("INCORRECT\n");
    }
    return 0;
}