Time Limit:1000MS Memory Limit:64000KB 64bit IO Format:%lld & %llu
Description
Sudoku is a popular single player game. The objective is to fill a 9x9 matrix with digits so that each column, each row, and all 9 non-overlapping 3x3 sub-matrices contain all of the digits from 1 through 9. Each 9x9 matrix is partially completed at the start of game play and typically has a unique solution.
Given a completed N2×N2 Sudoku matrix, your task is to determine whether it is a valid solution.
A valid solution must satisfy the following criteria:
- Each row contains each number from 1 to N2, once each.
- Each column contains each number from 1 to N2, once each.
- Divide the N2×N2 matrix into N2 non-overlapping N×N sub-matrices. Each sub-matrix contains each number from 1 to N2, once each.
You don't need to worry about the uniqueness of the problem. Just check if the given matrix is a valid solution.
Input
The first line of the input gives the number of test cases, T(1 ≤ T ≤ 100).
T test cases follow. Each test case starts with an integer N(3 ≤ N ≤ 6).
The next N2 lines describe a completed Sudoku solution, with each line contains exactly N2 integers.
All input integers are positive and less than 1000.
Output
For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is "Yes" (quotes for clarity only) if it is a valid solution, or "No" (quotes for clarity only) if it is invalid.
Sample Input
3
3
5 3 4 6 7 8 9 1 2
6 7 2 1 9 5 3 4 8
1 9 8 3 4 2 5 6 7
8 5 9 7 6 1 4 2 3
4 2 6 8 5 3 7 9 1
7 1 3 9 2 4 8 5 6
9 6 1 5 3 7 2 8 4
2 8 7 4 1 9 6 3 5
3 4 5 2 8 6 1 7 9
3
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
3
5 3 4 6 7 8 9 1 2
6 7 2 1 9 5 3 4 8
1 9 8 3 4 2 5 6 7
8 5 9 7 6 1 4 2 3
4 2 6 8 999 3 7 9 1
7 1 3 9 2 4 8 5 6
9 6 1 5 3 7 2 8 4
2 8 7 4 1 9 6 3 5
3 4 5 2 8 6 1 7 9
Sample Output
Case #1: Yes
Case #2: No
Case #3: No
题意：检查一个方阵是不是数独。
题解：技巧暴力。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=*;
int a[maxn][maxn];
bool vis[];//其实不用开这么大 超过n*n直接返回0了 开40就行 开大点保险
bool check(int n) //这里如果想不传参可以定义为全局变量
{
for(int i=; i<=n*n; i++)
{
memset(vis,,sizeof(vis));//注意初始化该放的位置
for(int j=; j<=n*n; j++)
{
if(a[i][j]<||a[i][j]>n*n) return ;//注意越界
if(vis[a[i][j]]) return ;
vis[a[i][j]]=;
}
}
for(int i=; i<=n*n; i++)
{
memset(vis,,sizeof(vis));
for(int j=; j<=n*n; j++)
{
if(a[j][i]<||a[j][i]>n*n) return ;//这里第一次直接复制的ij写反了 但是也过了 汗
if(vis[a[j][i]]) return ;
vis[a[j][i]]=;
}
}
for(int si=; si<n*n; si+=n)
for(int sj=; sj<n*n; sj+=n)
{
memset(vis,,sizeof(vis));
for(int i=; i<=n; i++)
{
for(int j=; j<=n; j++)
{
int x=si+i;
int y=sj+j;
if(a[x][y]<||a[x][y]>n*n) return ;
if(vis[a[x][y]]) return ;//注意vis是一维的
vis[a[x][y]]=;
}
}
}
return ;//默认也是返回1
}
int main()
{
int t,cas=,n;
cin>>t;
while(t--)
{
cin>>n;
for(int i=; i<=n*n; i++)
for(int j=; j<=n*n; j++)
cin>>a[i][j]; //也可以在这里判断一下如果越界直接No
if(check(n)) printf("Case #%d: Yes\n",cas++);//传n
else printf("Case #%d: No\n",cas++);
}
return ;
}