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P1074 靶形数独

 


 

按点排序太他喵玄学了,T两个点过不去,然后换行换列都不行,无奈按行排序,正常的搜索题


#include<bits/stdc++.h>
using namespace std;
int row[10][10],a,ln[10][10],num[10][10],ck[10][10],ans=-1,pt=0,tot,cg[10][10];
int vt[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}};
struct node{int i,j;}ab[1000];
struct nd{int tr,val;}rt[10];
int cmp(nd x,nd y){return x.val>y.val;}
void dfs(int pos)
{
        int i=ab[pos].i,j=ab[pos].j;
        int tk=cg[i][j];
        for(int k=1;k<=9;k++)
        if(!row[i][k]&&!ln[j][k]&&!ck[tk][k])
        {
            row[i][k]=ln[j][k]=ck[tk][k]=1;
            pt+=vt[i][j]*k;
            if(pos==tot) ans=max(ans,pt);
            else dfs(pos+1);
            row[i][k]=ln[j][k]=ck[tk][k]=0;
            pt-=vt[i][j]*k;
        }
}
int main()
{
    for(int i=1;i<=9;i++)
    for(int j=1;j<=9;j++)
    {    cg[i][j]=(i-1)/3*3+(j-1)/3+1;
        cin>>a;
        if(a)
        {
        row[i][a]=1;
        rt[i].val++;
        ln[j][a]=1;
        ck[cg[i][j]][a]=1;
        pt+=vt[i][j]*a;
        num[i][j]=a;
        }
    }
    for(int i=1;i<=9;i++)rt[i].tr=i;
    sort(rt+1,rt+10,cmp); 
    for(int i=1;i<=9;i++)
    {
        int ro=rt[i].tr;
        for(int j=1;j<=9;j++)
        if(!num[ro][j])ab[++tot].i=ro,ab[tot].j=j;
    }
    dfs(1);
    cout<<ans;
}

 

posted @ 2019-09-10 21:22  SFWR  Views(121)  Comments(0Edit  收藏  举报