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

卧槽，数独，判断行，列和小方格里的数不重复。但只判断当前状态。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
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;i<n*n;i++)
        {
            for(int j=0;j<n*n;j++)
            {
                scanf("%d",&a[i][j]);
                if(a[i][j]>0&&flag)
                {
                    if(!x[i][a[i][j]])//判断行
                       x[i][a[i][j]]=1;
                    else
                       flag=false;
                    if(!y[j][a[i][j]])//判断列
                       y[j][a[i][j]]=1;
                    else
                        flag=false;
                    if(!s[i/n*n+j/n+1][a[i][j]])//判断小方格
                        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;
}

POJ 3435 Sudoku Checker,布布扣,bubuko.com