技術標籤：DFS;BFS

>Link

ybtoj數獨遊戲

>解題思路

這一道題據zzl大佬所說跟八皇后問題很像（找了一下發現我當時沒有寫八皇后的部落格
為了使每一行每一列每一宮每一個數字只存在一個，可以建一個變數 h i , j h_{i,j} hi,j​表示第 i i i行是否存在數字 j j j，列和宮類似
然後通過所在區域是否存在當前數字進行剪枝

宮可以暴力進行判斷✔

>程式碼

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
 

using namespace std;

int a[100];
string s;
bool ans, h[20][20], l[20][20], g[20][20];

int getG (int x)
{
	int H = x / 9 + 1;
	int L = x % 9;
	if (x % 9 == 0) H--, L = 9;
	if (H <= 3)
	{
		if (L <= 3) return 1;
		if (L <= 6) return 2;
		if (L <= 9) return 3;
	}
	if (H <= 6)
	{
		if (L <= 3
 
) return 4;
		if (L <= 6) return 5;
		if (L <= 9) return 6;
	}
	if (L <= 3) return 7;
	if (L <= 6) return 8;
	return 9;
}
void dfs (int now)
{
	if (ans) return;
	if (now == 81 + 1)
	{
		ans = 1;
		for (int i = 1; i <= 81; i++) printf ("%d", a[i]);
		printf ("\n");
		return
 
;
	}
	if (a[now] != 0) {dfs (now + 1); return;} 
	int H = now / 9 + 1;
	int L = now % 9;
	if (now % 9 == 0) H--, L = 9;
	int G = getG (now);
	for (int i = 1; i <= 9; i++)
	  if (!h[H][i] && !l[L][i] && !g[G][i])
	  {
	  	a[now] = i;
		h[H][i] = l[L][i] = g[G][i] = 1;
		dfs (now + 1);
		a[now] = 0;
		h[H][i] = l[L][i] = g[G][i] = 0;
	  }
}

int main()
{
	cin >> s;
	while (s != "end")
	{
		ans = 0;
		memset (a, 0, sizeof (a));
		memset (h, 0, sizeof (h));
		memset (l, 0, sizeof (l));
		memset (g, 0, sizeof (g));
		for (int i = 1; i <= 81; i++)
		  if (s[i - 1] != '.')
		  {
		  	a[i] = s[i - 1] - '0';
			int H = i / 9 + 1;
			int L = i % 9;
			if (i % 9 == 0) H--, L = 9;
		  	h[H][a[i]] = l[L][a[i]] = g[getG (i)][a[i]] = 1;
		  }
		dfs (1);
		cin >> s;
	}
	return 0;
}