Acdream semi-finals guide series (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 completedN2 × N2Sudoku matrix, your task is to determine whether it is a valid solution.

A valid solution must satisfy the following criteria:

• Each row contains each numberFrom 1 to N2, Once each.
• Each column contains each numberFrom 1 to N2, Once each.
• DivideN2 × N2MatrixN2Non-overlappingN × nSub-matrices. Each sub-matrix contains each numberFrom 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).

TTest Cases Follow. Each test case starts with an integerN (3 ≤ n ≤ 6).

The nextN2Lines describe a completed Sudoku solution, with each line contains exactlyN2Integers.

All input integers are positive and less1000.

Output

For each test case, output one line containing "case # X: Y", whereXIs the case number (starting from 1) andYIs"Yes"(Quotes for clarity only) if it is a valid solution, or"No"(Quotes for clarity only) if it is invalid.

Sample Input

335 3 4 6 7 8 9 1 26 7 2 1 9 5 3 4 81 9 8 3 4 2 5 6 78 5 9 7 6 1 4 2 34 2 6 8 5 3 7 9 17 1 3 9 2 4 8 5 69 6 1 5 3 7 2 8 42 8 7 4 1 9 6 3 53 4 5 2 8 6 1 7 931 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 935 3 4 6 7 8 9 1 26 7 2 1 9 5 3 4 81 9 8 3 4 2 5 6 78 5 9 7 6 1 4 2 34 2 6 8 999 3 7 9 17 1 3 9 2 4 8 5 69 6 1 5 3 7 2 8 42 8 7 4 1 9 6 3 53 4 5 2 8 6 1 7 9

Sample output

Case #1: YesCase #2: NoCase #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; // the rule here is very important !! Cin> A; if (a <1 | A> N * n) {flag = true; continue ;} // check each row if (row [I] [a]) {flag = true;} else row [I] [a] = true; // check each column if (COL [J] [a]) {flag = true;} else Col [J] [a] = true; // check each small rectangle. 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 semi-finals guide series (6)