洛谷 1784 数独（hdu 1426 Sudoku Killer）

洛谷1784 数独：
首先题意是，给你一个9x9的未填满的数独（未填满用0表示），要求你将这些0的位置上填上数字，以满足数独的特性。
解析：
其实这道题爆搜就行，毕竟只是9x9的数独，直接从每个0的位置开始搜即可
上代码：

#include <bits/stdc++.h>
#define ll long long

using namespace std;

typedef pair<int, int> pii;
const int N=9+5;
int x[N*N],y[N*N];
int mp[N][N];
bool ok = 0;
int cnt;

bool check(int dx,int dy)
{
	for (int i=1; i<=9; i++)if (i!=dx&&mp[i][dy]==mp[dx][dy])return false;
	for (int i=1; i<=9; i++)if (i!=dy&&mp[dx][i]==mp[dx][dy])return false;
	int basex=dx;
	int basey=dy;
	basex--;
	basex/=3;
	basex*=3;
	basex++;
	basey--;
	basey/=3;
	basey*=3;
	basey++;
	for (int i=0; i<=2; i++)
		for (int j=0; j<=2; j++)
		{
			if (((i+basex)!=dx)&&(j+basey!=dy))
				if (mp[i+basex][basey+j]==mp[dx][dy])
					return false;
		}
	return true;
}

void dfs(int now)
{
	if (ok) return;
	if (now == cnt)
	{
		for (int i = 1; i <= 9; i++)
		{
			for (int j = 1; j <= 9; j++)
			{
				printf("%d%c", mp[i][j], j == 9 ? '\n' : ' ');
			}
		}
		ok = 1;
		return;
	}
	for (int i = 1; i <= 9; i++)
	{
		mp[x[now]][y[now]] = i;
		if (check(x[now],y[now]))dfs(now + 1);
		mp[x[now]][y[now]] = 0;
	}
	return;
}
int main()
{
		for (int i=1; i<=9; i++)
		{
			for (int j=1; j<=9; j++)
			{
				int now;
				scanf("%d",&now);
				if (now!=0)
					mp[i][j]=now;
				else
				{
					mp[i][j]=0;
					x[cnt] = i;
					y[cnt++] = j;
				}
			}
		}
		dfs(0);
	return 0;
}

hdu1426
这道题其实是与上题是差不多的，只不过是将0换成了问号，还有随便吐槽一下hdu的辣鸡评测系统，PE改的快要疯了，洛谷最好了

#include <bits/stdc++.h>
#define ll long long

using namespace std;

typedef pair<int, int> pii;
const int N=9+5;
int x[N*N],y[N*N];
int mp[N][N];
bool ok = 0;
int _cnt=0;
int cnt;

bool check(int dx,int dy)
{
	for (int i=1; i<=9; i++)if (i!=dx&&mp[i][dy]==mp[dx][dy])return false;
	for (int i=1; i<=9; i++)if (i!=dy&&mp[dx][i]==mp[dx][dy])return false;
	int basex=dx;
	int basey=dy;
	basex--;
	basex/=3;
	basex*=3;
	basex++;
	basey--;
	basey/=3;
	basey*=3;
	basey++;
	for (int i=0; i<=2; i++)
		for (int j=0; j<=2; j++)
		{
			if (((i+basex)!=dx)&&(j+basey!=dy))
				if (mp[i+basex][basey+j]==mp[dx][dy])
					return false;
		}
	return true;
}

void dfs(int now)
{
	if (ok) return;
	if (now == cnt)
	{
		for (int i = 1; i <= 9; i++)
		{
			for (int j = 1; j <= 9; j++)
			{
				printf("%d%c", mp[i][j], j == 9 ? '\n' : ' ');
			}
		}
		ok = 1;
		return;
	}
	for (int i = 1; i <= 9; i++)
	{
		mp[x[now]][y[now]] = i;
		if (check(x[now],y[now]))dfs(now + 1);
		mp[x[now]][y[now]] = 0;
	}
	return;
}
int main()
{
	char ch;
	while (cin >> ch)
	{
		ok=0;
		cnt = 0;
		if (_cnt)
			printf("\n");
		_cnt++;
		if (isdigit(ch))
			mp[1][1]=ch-'0';
		else
		{
			mp[1][1]=0;
			x[cnt] = 1;
			y[cnt++] = 1;
		}
		for (int i=1; i<=9; i++)
		{
			for (int j=1; j<=9; j++)
			{
				if (i==1&&j==1)continue;

				cin>>ch;
				if (isdigit(ch))
					mp[i][j]=ch-'0';
				else
				{
					mp[i][j]=0;
					x[cnt] = i;
					y[cnt++] = j;
				}
			}
		}
		dfs(0);
	}
	return 0;
}