sudoku1
7 0 0 9 0 0 0 0 1
1 0 0 0 0 5 9 0 0
0 0 0 2 0 0 0 8 0
0 0 5 0 2 0 0 0 3
0 0 0 0 0 0 6 4 8
4 1 3 0 0 0 0 0 0
0 0 7 0 0 2 0 9 0
2 0 1 0 6 0 8 0 4
0 8 0 5 0 4 0 1 2
sudoku2
0 0 0 7 0 2 4 5 3
9 0 0 0 0 8 0 0 0
7 4 0 0 0 5 0 1 0
1 9 5 0 8 0 0 0 0
0 7 0 0 0 0 0 2 5
0 3 0 5 7 9 1 0 8
0 0 0 6 0 1 0 0 0
0 6 0 9 0 0 0 0 1
0 0 0 0 0 0 0 0 6
sudoku1
2829
sudoku2
2852
以后学习位运算的做法及dancing links
#include<iostream>
#include<cstdio>
using namespace std;
int x,t=81,ans,flag;
int a[10][10],row[10][10],col[10][10],block[10][10];
int v[10][10]=
{
0,0,0,0,0, 0,0,0,0,0,
0,6,6,6,6, 6,6,6,6,6,
0,6,7,7,7, 7,7,7,7,6,
0,6,7,8,8, 8,8,8,7,6,
0,6,7,8,9, 9,9,8,7,6,
0,6,7,8,9,10,9,8,7,6,
0,6,7,8,9, 9,9,8,7,6,
0,6,7,8,8, 8,8,8,7,6,
0,6,7,7,7, 7,7,7,7,6,
0,6,6,6,6, 6,6,6,6,6
};
int ju[10][10]={
0,0,0,0,0,0,0,0,0,0,
0,1,1,1,2,2,2,3,3,3,
0,1,1,1,2,2,2,3,3,3,
0,1,1,1,2,2,2,3,3,3,
0,4,4,4,5,5,5,6,6,6,
0,4,4,4,5,5,5,6,6,6,
0,4,4,4,5,5,5,6,6,6,
0,7,7,7,8,8,8,9,9,9,
0,7,7,7,8,8,8,9,9,9,
0,7,7,7,8,8,8,9,9,9,
};
//void print()
//{
// for(int i=1;i<=9;i++)
// {
// for(int j=1;j<=9;j++)
// cout<<a[i][j]<<" ";
// cout<<endl;
// }
// cout<<endl;
//}
void dfs(int rm,int now) {
if(rm==0) {flag=1; ans=max(ans,now); return;}
if(now+rm*90<=ans) return;
int tx=0,ty=0,mn=10,way=0,bc=0;
for(register int i=1;i<=9;i++) //81格当中放那个数的方法最少找出来
for(register int j=1;j<=9;j++)
if(!a[i][j])
{
way=9,bc=ju[i][j];
for(register int k=1;k<=9;k++)
{
if(row[i][k] || col[j][k] || block[bc][k]) way--;
}
if(way<=mn) mn=way,tx=i,ty=j;
}
int tc=ju[tx][ty];
for(register int i=1;i<=9;i++) //先放置可选择数目最少的空位,如果在上面可以记录可以放那些数更好
if(!row[tx][i] && !col[ty][i] && !block[tc][i])
{
a[tx][ty]=i;
row[tx][i]=1;
col[ty][i]=1; //row[][]...err
block[tc][i]=1;
//
// cout<<tx<<" "<<ty<<endl;
// print();
//
dfs(rm-1,now+a[tx][ty]*v[tx][ty]);
row[tx][i]=0;
col[ty][i]=0;
block[tc][i]=0;
a[tx][ty]=0;
}
}
int main()
{
// freopen("shudu.txt","r",stdin);
for(int i=1;i<=9;i++)
for(int j=1;j<=9;j++)
{
scanf("%d",&a[i][j]);
if(a[i][j]>0)
{
ans+=a[i][j]*v[i][j];
row[i][a[i][j]]=1;
col[j][a[i][j]]=1;
block[ju[i][j]][a[i][j]]=1;
t--;
}
}
dfs(t,ans);
if(flag) printf("%d\n",ans);
else printf("-1\n");
// cout<<clock()<<endl;
return 0;
}