3109. [CQOI2013]新数独搜索

Posted 2020-10-29 Refun

Description

Input

输入一共15行，包含一个新数独的实例。第奇数行包含左右方向的符号（<和>），第偶数行包含上下方向的符号（^和v）。

 

Output

输出包含9行，每行9个1~9的数字，以单个空格隔开。输入保证解惟一。

Sample Input

< > > < > <
v v ^ ^ v v ^ ^ ^
< < > < > <
^ ^ ^ v ^ ^ ^ v v
< < < < > >
> < > > > >
v ^ ^ ^ ^ v v v ^
> > > > < >
v v ^ v ^ v ^ v ^
> < < > > >
< < < < > <
v ^ v v v v ^ ^ v
< > > < < >
^ v v v ^ v ^ v v
< > < > < >

Sample Output

4 9 1 7 3 6 5 2 8
2 3 7 8 1 5 6 4 9
5 6 8 2 4 9 7 3 1
9 1 3 6 5 4 8 7 2
8 5 4 9 7 2 1 6 3
7 2 6 3 8 1 9 5 4
3 4 9 5 6 8 2 1 7
1 8 5 4 2 7 3 9 6
6 7 2 1 9 3 4 8 5

 

比靶形数独那个题水多了= =……

 

#include<iostream>
#include<cstring>
#include<cstdio>
#define id(x,y) (x-1)*9+y
using namespace std;
int Relation[101][101],ans[12][12],maxn;
int dx[4]= {0,0,-1},dy[4]= {0,-1,0};
bool row[10][10],column[10][10],square[10][10][10],flag;
char st[101];

void Dfs(int x,int y)
{
	maxn=max(x,maxn);
	if (x==10 && y==1)
	{
		flag=true;
		for (int i=1; i<=9; ++i)
		{
			for (int j=1; j<=8; ++j)
				printf("%d ",ans[i][j]);
			printf("%d\n",ans[i][9]);
		}
		return;
	}
	int down=1,up=9;
	for (int i=1; i<=2; ++i)
	{
		int xx=x+dx[i],yy=y+dy[i];
		if (xx<1 || xx>9 || yy<1 || yy>9 || ans[xx][yy]==0) continue;
		if ( Relation[id(x,y)][id(xx,yy)] == 1 ) down=max(down,ans[xx][yy]+1);
		if ( Relation[id(x,y)][id(xx,yy)] == 0 ) up=min(up,ans[xx][yy]-1);
	}
	for (int i=down; i<=up; ++i)
		if (!row[x][i] && !column[y][i] && !square[(x-1)/3][(y-1)/3][i])
		{
			ans[x][y]=i;
			row[x][i]=column[y][i]=square[(x-1)/3][(y-1)/3][i]=true;
			if (y==9) Dfs(x+1,1);
			else Dfs(x,y+1);
			ans[x][y]=0;
			row[x][i]=column[y][i]=square[(x-1)/3][(y-1)/3][i]=false;

			if (flag) return;
		}
}

int main()
{
	memset(Relation,-1,sizeof(Relation));
	for (int i=1; i<=15; ++i)
		if (i%2==(1^(i>=6 && i<=10)))
		{
			for (int j=1; j<=3; ++j)
				for (int k=1; k<=2; ++k)
				{
					char opt=getchar();
					while (opt!=‘>‘ && opt!=‘<‘) opt=getchar();
					if (opt==‘>‘)
					{
						Relation[(j-1)*3+k+(i/2+(i>10))*9][(j-1)*3+k+(i/2+(i>10))*9+1]=1;
						Relation[(j-1)*3+k+(i/2+(i>10))*9+1][(j-1)*3+k+(i/2+(i>10))*9]=0;
					}
					else
					{
						Relation[(j-1)*3+k+(i/2+(i>10))*9+1][(j-1)*3+k+(i/2+(i>10))*9]=1;
						Relation[(j-1)*3+k+(i/2+(i>10))*9][(j-1)*3+k+(i/2+(i>10))*9+1]=0;
					}
				}

		}
		else
		{
			for (int j=1; j<=9; ++j)
			{
				char opt=getchar();
				while (opt!=‘^‘ && opt!=‘v‘) opt=getchar();
				if (opt==‘v‘)
				{
					Relation[id(i/2+(i>=5),j)][id(i/2+(i>=5)+1,j)]=1;
					Relation[id(i/2+(i>=5)+1,j)][id(i/2+(i>=5),j)]=0;
				}
				else
				{
					Relation[id(i/2+(i>=5)+1,j)][id(i/2+(i>=5),j)]=1;
					Relation[id(i/2+(i>=5),j)][id(i/2+(i>=5)+1,j)]=0;
				}
			}
		}
	Dfs(1,1);
}