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ACdream区域赛指导赛之手速赛系列(6)
Problem 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-overlappingN×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 exactlyN2 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) andy 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
#include<iostream> using namespace std; int n; int small[40][40]; int col[40][40]; int row[40][40]; int main() { int T; cin>>T; int a; int k; for(int count = 0;count<T;count++) { bool flag = false; cin>>n; memset(small,0,sizeof(small)); memset(col,0,sizeof(col)); memset(row,0,sizeof(row)); for(int i=0;i<n*n;i++){ for(int j=0;j<n*n;j++){ k = (i/n)*n+j/n; //这里的规律总结出来很重要!! cin>>a; if(a<1||a>n*n){ flag=true; continue; } // 对每行进行检查 if(row[i][a]){ flag=true; } else row[i][a]=true; // 对每列进行检查 if(col[j][a]){ flag=true; } else col[j][a]=true; // 对每个小的矩形块进行检查 if(small[k][a]){ flag=true; } else { small[k][a]=true; } } } if(!flag)printf("Case #%d: Yes\n",count); else printf("Case #%d: No\n",count); } return 0; }
ACdream区域赛指导赛之手速赛系列(6)