Problem J. Let Sudoku Rotate

Problem Description

Sudoku is a logic-based, combinatorial number-placement puzzle, which is popular around the world.
In this problem, let us focus on puzzles with 16×16 grids, which consist of 4×4 regions. The objective is to fill the whole grid with hexadecimal digits, i.e. 0123456789ABCDEF, so that each column, each row, and each region contains all hexadecimal digits. The figure below shows a solved sudoku.
Yesterday, Kazari solved a sudoku and left it on the desk. However, Minato played a joke with her - he performed the following operation several times.
* Choose a region and rotate it by 90 degrees counterclockwise.
She burst into tears as soon as she found the sudoku was broken because of rotations.
Could you let her know how many operations her brother performed at least?

Input

The first line of the input contains an integer T (1≤T≤103) denoting the number of test cases.
Each test case consists of exactly 16 lines with 16 characters each, describing a broken sudoku.

Output

For each test case, print a non-negative integer indicating the minimum possible number of operations.

暴力dfs,對每個矩陣的旋轉次數進行列舉。

搜尋過程當中，可以利用數獨的性質剪枝，也可以利用迭代加深，噹噹前花費多於答案時直接return.

#include <cstdio>
#include <iostream>
#include <string.h>
#include <string> 
#include <map>
#include <queue>
#include <deque>
#include <vector>
#include <set>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <assert.h>
#define pb push_back 
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
using namespace std;
typedef long long ll;
typedef long double ld;
typedef double db;
typedef pair<int,int> pp;
const int maxn=25,inf=0x3f3f3f3f;
const ll llinf=0x3f3f3f3f3f3f3f3f;
int r[maxn][maxn],c[maxn][maxn];
int x11[16]={1,1,1,1,5,5,5,5,9,9,9,9,13,13,13,13},
    x22[16]={4,4,4,4,8,8,8,8,12,12,12,12,16,16,16,16},
    y11[16]={1,5,9,13,1,5,9,13,1,5,9,13,1,5,9,13},
    y22[16]={4,8,12,16,4,8,12,16,4,8,12,16,4,8,12,16};
int a[maxn][maxn][4];
char s[maxn][maxn];
int ans;

void rotate(int id,int re) {
/*    for (int i=x11[id];i<=x22[id];i++) {
        for (int j=y11[id];j<=y22[id];j++) {
            cout << a[i][j][re] << ' ';
        }
        cout << '\n';
    }*/
    for (int i=x11[id];i<=x22[id];i++) {
        for (int j=y11[id];j<=y22[id];j++) {
//            a[x22[id]-(j-y11[id])][i-x11[id]+y11[id]][re+1]=a[i][j][re];
            a[j-y11[id]+x11[id]][x22[id]-i+y11[id]][re+1]=a[i][j][re];
        }
    }
/*    for (int i=x11[id];i<=x22[id];i++) {
        for (int j=y11[id];j<=y22[id];j++) {
            cout << a[i][j][re+1] << ' ';
        }
        cout << '\n';
    }*/
}

void dfs(int id,int cost) {
    if (id==16) {
        ans=min(ans,cost);
        return;
    }
    for (int k=0;k<4;k++) {
        int flag=1;
        if (cost+k>ans) continue;
        for (int i=x11[id];i<=x22[id];i++) {
            for (int j=y11[id];j<=y22[id];j++) {
                if (r[i][a[i][j][k]]) {
                    flag=0;break;
                }
                if (c[j][a[i][j][k]]) {
                    flag=0;break;
                }
            }
        }
        if (flag) {
            for (int i=x11[id];i<=x22[id];i++) 
                for (int j=y11[id];j<=y22[id];j++) 
                    r[i][a[i][j][k]]=c[j][a[i][j][k]]=1;
            dfs(id+1,cost+k);
            for (int i=x11[id];i<=x22[id];i++) 
                for (int j=y11[id];j<=y22[id];j++) 
                    r[i][a[i][j][k]]=c[j][a[i][j][k]]=0;
        }
    }
}

int main() {
    int cas;
    scanf("%d",&cas);
    while (cas--) {
        for (int i=1;i<=16;i++) {
            scanf("%s",s[i]+1);
            for (int j=1;j<=16;j++) {
                if (s[i][j]>='A'&&s[i][j]<='Z') {
                    a[i][j][0]=s[i][j]-'A'+10;
                } else a[i][j][0]=s[i][j]-'0';
            }
        }
        for (int j=0;j<=2;j++) {
            for (int k=0;k<16;k++) 
                rotate(k,j);
        }
    /*    for (int k=0;k<4;k++){
        for (int i=1;i<=16;i++) {
            for (int j=1;j<=16;j++) {
                cout << a[i][j][k] << ' ';
                if (a[i][j][k]<=9) cout << ' '; 
            }
            cout << '\n';
        }
        cout << '\n';}*/
        ans=inf;
        mem0(r);mem0(c);
        dfs(0,0);
        assert(ans!=inf);
        printf("%d\n",ans);
    }
    return 0;
}