P1074 靶状数独（优化）

题见洛谷

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#define LL long long
using namespace std;
bool lief[10][10],hangf[10][10],gef[10][10];
int ans=-1,can[10][10],n=0,a[10][10];//can[][]用来记此点能放几种数
int quan[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 ge[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,
};
bool flag(int x,int y,int k)
{
    if(!gef[ge[x][y]][k]&&!hangf[x][k]&&!lief[y][k]) return true;

    return false;
}
void dfs(int k)
{

    if(k>n)
    {
        int tot=0;
        for(int i=1;i<=9;i++)
         for(int j=1;j<=9;j++)
         if(!a[i][j])
             return;
         else
             tot+=a[i][j]*quan[i][j];
        ans=max(ans,tot);
        return;
    }

    else
    {
        int minn=99999999,nx,ny;
        memset(can,0,sizeof(can));
        for(int i=1;i<=9;i++)
         for(int j=1;j<=9;j++)
          if(!a[i][j])
          {
            for(int x=1;x<=9;x++)
            if(flag(i,j,x)) can[i][j]++;
            if(minn>can[i][j])
            {
                minn=can[i][j];
                nx=i;ny=j;//找到最小can的，先来填
            } 
            if(minn==1)break;

          }

        if(minn==0)return;//无解
        if(minn==99999999)      
        {
            int tot=0;
            for(int i=1;i<=9;i++)
             for(int j=1;j<=9;j++)
             if(!a[i][j])
                return;
             else
                tot+=a[i][j]*quan[i][j];
            ans=max(ans,tot);
            return;
        }

        for(int i=1;i<=9;i++)
         if(flag(nx,ny,i))
        {
            hangf[nx][i]=true;
            lief[ny][i]=true;
            gef[ge[nx][ny]][i]=true;
            a[nx][ny]=i;

            dfs(k+1);

            hangf[nx][i]=false;
            lief[ny][i]=false;
            gef[ge[nx][ny]][i]=false;
            a[nx][ny]=0;
        }
    }
}
int main()
{
    for(int i=1;i<=9;i++)
     for(int j=1;j<=9;j++)
    {
        scanf("%d",&a[i][j]);
        if(!a[i][j])n++;
        else
        {
            lief[j][a[i][j]]=true;
            hangf[i][a[i][j]]=true;
            gef[ge[i][j]][a[i][j]]=true;
        }


    }
    dfs(1);
    printf("%d",ans);
    return 0;
}