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 longusing 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;}