【ybt 入门到进阶模拟赛 Day2 T4】数独游戏

数独游戏

题目链接：ybt 入门到进阶模拟赛 Day2 T4

题目大意

就要你填一个数独，保证有唯一解。

思路

直接暴力 dfs，不用剪枝优化都可以过。

代码

#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

struct zb {
	int x, y;
};
char a[11][11];
bool can, used[11][11][11];
int know[11][11];
queue <zb> q;
int heng[11][11], zong[11][11], blogs[11][11];

int get_kuai(int x, int y) {//求属于哪个小区间
	if (x <= 3 && y <= 3) return 1;
	if (x <= 3 && y <= 6) return 2;
	if (x <= 3 && y <= 9) return 3;
	if (x <= 6 && y <= 3) return 4;
	if (x <= 6 && y <= 6) return 5;
	if (x <= 6 && y <= 9) return 6;
	if (x <= 9 && y <= 3) return 7;
	if (x <= 9 && y <= 6) return 8;
	if (x <= 9 && y <= 9) return 9;
}

void dfs(int nowx, int nowy) {
	if (nowx > 9) {
		can = 1;
		return ;
	}
	if (can) return ;
	
	if (a[nowx][nowy] != '?') {
		if (nowy == 9) {
			dfs(nowx + 1, 1);
		}
		else dfs(nowx, nowy + 1);
		return ;
	}
	
	for (int i = 1; i <= 9; i++)
		if (!heng[nowx][i] && !zong[nowy][i] && !blogs[get_kuai(nowx, nowy)][i]) {//不会有冲突
			heng[nowx][i] = 1;
			zong[nowy][i] = 1;
			blogs[get_kuai(nowx, nowy)][i] = 1;
			a[nowx][nowy] = i + '0';
			if (nowy == 9) {
				dfs(nowx + 1, 1);
			}
			else dfs(nowx, nowy + 1);
			if (can) return ;
			heng[nowx][i] = 0;
			zong[nowy][i] = 0;
			blogs[get_kuai(nowx, nowy)][i] = 0;
			a[nowx][nowy] = '?';
		}
}

int main() {
	freopen("sudoku.in", "r", stdin);
	freopen("sudoku.out", "w", stdout);
	
	for (int i = 1; i <= 9; i++) {
		for (int j = 1; j <= 9; j++) {
			a[i][j] = getchar();
			while (a[i][j] != '?' && !(a[i][j] >= '1' && a[i][j] <= '9'))
				a[i][j] = getchar();
			if (a[i][j] != '?') {
				heng[i][a[i][j] - '0'] = 1;
				zong[j][a[i][j] - '0'] = 1;
				blogs[get_kuai(i, j)][a[i][j] - '0'] = 1;
			}
		}
	}
	
	dfs(1, 1);
	
	for (int i = 1; i <= 9; i++) {
		for (int j = 1; j <= 9; j++) {
			printf("%c", a[i][j]);
		}
		printf("\n");
	}
	
	return 0;
}