CSU 1605 数独

 题目大意：

9宫格每个位置都有对应的分数，填完数独后根据对应位置的分数相加之和求个最大值，不存在输出-1

说什么用位运算加速可以解决问题，但是对着标程还是T，最近学了dlx，发现这样解决数独快了很多

位运算加速我确实写不出了，直接用dlx来做这道题目

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <climits>
#include <cmath>

using namespace std;
#define N 1000
#define MAXNODE 1000000
const int INF = INT_MAX;
const double eps = 1e-8;

int a[10][10];
char str[5];

int belong[9][9] =  {
                        {1,1,1,2,2,2,3,3,3},
                        {1,1,1,2,2,2,3,3,3},
                        {1,1,1,2,2,2,3,3,3},
                        {4,4,4,5,5,5,6,6,6},
                        {4,4,4,5,5,5,6,6,6},
                        {4,4,4,5,5,5,6,6,6},
                        {7,7,7,8,8,8,9,9,9},
                        {7,7,7,8,8,8,9,9,9},
                        {7,7,7,8,8,8,9,9,9},
                    };
int sc[9][9] = {
                    {6,6,6,6,6,6,6,6,6},
                    {6,7,7,7,7,7,7,7,6},
                    {6,7,8,8,8,8,8,7,6},
                    {6,7,8,9,9,9,8,7,6},
                    {6,7,8,9,10,9,8,7,6},
                    {6,7,8,9,9,9,8,7,6},
                    {6,7,8,8,8,8,8,7,6},
                    {6,7,7,7,7,7,7,7,6},
                    {6,6,6,6,6,6,6,6,6},
               };


void printM()
{
    for(int i=0 ; i<9 ; i++)
        for(int j=0 ; j<9 ; j++){
           if(j<8) printf("%d " , a[i][j]);
           else printf("%d
" , a[i][j]);
        }
}

struct DLX{
    int n ,m , size;
    int col[MAXNODE] , row[MAXNODE];
    int U[MAXNODE] , D[MAXNODE] , L[MAXNODE] , R[MAXNODE];
    int cnt_col[N] , first[N];
    int ans[100] , minv;

    void init(int _n , int _m)
    {
        n = _n , m = _m;
        size= m ;
        for(int i=0 ; i<=m ; i++){
            L[i] = i-1 , R[i] = i+1;
            U[i] = D[i] = i;
        }
        L[0] = m , R[m] = 0;
        for(int i=1 ; i<=m ; i++) cnt_col[i] = 0;
        for(int i=1 ; i<=n ; i++) first[i] = -1;
        minv = 0;
    }

    void link(int r , int c)
    {
        ++size;
        U[D[c]] = size , D[size] = D[c];
        U[size] = c , D[c] = size;

        if(first[r]<0) L[size]=R[size]=first[r] = size;
        else{
            L[R[first[r]]] = size , R[size] = R[first[r]];
            L[size] = first[r] , R[first[r]] = size;
        }
        row[size] = r , col[size] = c , cnt_col[c]++;
    }

    void Remove(int c)
    {
        L[R[c]] = L[c] , R[L[c]] = R[c];
        for(int i=D[c] ; i!=c ; i=D[i]){
            for(int j=R[i] ; j!=i ; j=R[j]){
                U[D[j]] = U[j] , D[U[j]] = D[j];
                cnt_col[col[j]]--;
            }
        }
    }

    void Resume(int c)
    {
        for(int i=U[c] ; i!=c ; i=U[i]){
            for(int j=L[i] ; j!=i ; j=L[j]){
                U[D[j]] = D[U[j]] = j;
                cnt_col[col[j]]++;
            }
        }
      //  printM();
        L[R[c]] = R[L[c]] = c;
    }

    void Dance(int d)
    {
        if(!R[0]){
            int v = 0;
            for(int i=0 ; i<d ; i++){
                    int r = (ans[i]-1)/81;
                    int c = ((ans[i]-1)%81)/9;
                    a[r][c] = ((ans[i]-1)%9)+1;
                    v+=a[r][c]*sc[r][c];
            }
            minv=max(minv , v);
            return;
        }
        int st=R[0];
        for(int i=R[0] ; i!=0 ; i=R[i])
            if(cnt_col[i]<cnt_col[st])
                st = i;
        Remove(st);
        for(int i=D[st] ; i!=st ; i=D[i]){
            ans[d] = row[i];
            for(int j=R[i] ; j!=i ; j=R[j]) Remove(col[j]);
            Dance(d+1);
            for(int j=L[i] ; j!=i ; j=L[j]) Resume(col[j]);
        }
        Resume(st);
        return ;
    }

}dlx;


int main()
{
  //  freopen("a.in" , "r" , stdin);
    int T;
    scanf("%d" , &T);
    while(T--)
    {
        for(int i=0 ; i<9 ; i++){
            for(int j=0 ; j<9 ; j++){
                scanf("%d" , &a[i][j]);
            }
        }
        dlx.init(729 , 324);
        for(int i=0 ; i<9 ; i++)
            for(int j=0 ; j<9 ; j++)
            {
                if(a[i][j]){
                    dlx.link((i*9+j)*9+a[i][j] , i*9+a[i][j]);
                    dlx.link((i*9+j)*9+a[i][j] , 81+j*9+a[i][j]);
                    dlx.link((i*9+j)*9+a[i][j] , 162+(belong[i][j]-1)*9+a[i][j]);
                    dlx.link((i*9+j)*9+a[i][j] , 243+i*9+j+1);
                }
                else{
                    for(int k=1 ; k<=9 ; k++){
                        dlx.link((i*9+j)*9+k , i*9+k);
                        dlx.link((i*9+j)*9+k , 81+j*9+k);
                        dlx.link((i*9+j)*9+k , 162+(belong[i][j]-1)*9+k);
                        dlx.link((i*9+j)*9+k , 243+i*9+j+1);
                    }
                }
            }
        dlx.Dance(0);
        if(!dlx.minv) puts("-1");
        else printf("%d
" , dlx.minv);
    }

    return 0;
}