POJ 2676

9阶数独，如上题做

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;

int puzzle[10][10];
char tmp[15];
const int SLOT=0;
const int ROW=1;
const int COL=2;
const int SUB=3;
const int maxn=400;
const int maxnode=9*9*9*9*9*4+100;
const int maxr=9*9*9*9*9*4+100;

struct DLX
{
  int n , sz;                                                 // 行数，节点总数
  int S[maxn];                                                // 各列节点总数
  int row[maxnode],col[maxnode];                              // 各节点行列编号
  int L[maxnode],R[maxnode],U[maxnode],D[maxnode];            // 十字链表

  int ansd,ans[maxn];                                         // 解

  void init(int n )
  {
    this->n = n ;
    for(int i = 0 ; i <= n; i++ )
    {
      U[i] = i ;
      D[i] = i ;
      L[i] = i - 1;
      R[i] = i + 1;
    }
    R[n] = 0 ;
    L[0] = n;
    sz = n + 1 ;
    memset(S,0,sizeof(S));
  }
  void addRow(int r,vector<int> c1)
  {
    int first = sz;
    for(int i = 0 ; i < c1.size(); i++ ){
      int c = c1[i];
      L[sz] = sz - 1 ; R[sz] = sz + 1 ; D[sz] = c ; U[sz] = U[c];
      D[U[c]] = sz; U[c] = sz;
      row[sz] = r; col[sz] = c;
      S[c] ++ ; sz ++ ;
    }
    R[sz - 1] = first ; L[first] = sz - 1;
  }
  // 顺着链表A，遍历除s外的其他元素
  #define FOR(i,A,s) for(int i = A[s]; i != s ; i = A[i])

  void remove(int c){
    L[R[c]] = L[c];
    R[L[c]] = R[c];
    FOR(i,D,c)
      FOR(j,R,i) {U[D[j]] = U[j];D[U[j]] = D[j];--S[col[j]];}
  }
  void restore(int c){
    FOR(i,U,c)
      FOR(j,L,i) {++S[col[j]];U[D[j]] = j;D[U[j]] = j; }
    L[R[c]] = c;
    R[L[c]] = c;
  }
  bool dfs(int d){
    if(R[0] == 0 ){
      ansd = d;
      return true;
    }
    // 找S最小的列c
    int c = R[0] ;
    FOR(i,R,0) if(S[i] < S[c]) c = i;

    remove(c);
    FOR(i,D,c){
      ans[d] = row[i];
      FOR(j,R,i) remove(col[j]);
      if(dfs(d + 1)) return true;
      FOR(j,L,i) restore(col[j]);
    }
    restore(c);

    return false;
  }
  bool solve(vector<int> & v){
    v.clear();
    if(!dfs(0)) return false;
    for(int i = 0 ; i< ansd ;i ++ ) v.push_back(ans[i]);
    return true;
  }
};
DLX slover;

int encode(int a,int b,int c){
	return a*81+b*9+c+1;
}

void decode(int code,int &a,int &b,int &c){
	code--;
	c=code%9; code/=9;
	b=code%9; code/=9;
	a=code;
}

void read(){
	for(int i=0;i<9;i++){
		scanf("%s",tmp);
		for(int j=0;j<9;j++)
		puzzle[i][j]=tmp[j]-'0';
	}
}

int main(){
	int kase;
	scanf("%d",&kase);
	while(kase--){
		read();
		slover.init(324);
		for(int r=0;r<9;r++){
			for(int c=0;c<9;c++){
				for(int v=0;v<9;v++){
					if(puzzle[r][c]==0||puzzle[r][c]==v+1){
						vector<int> columns;
						columns.push_back(encode(SLOT,r,c));
						columns.push_back(encode(ROW,r,v));
						columns.push_back(encode(COL,c,v));
						columns.push_back(encode(SUB,(r/3)*3+c/3,v));
						slover.addRow(encode(r,c,v),columns);
					}
				}
			}
		}
		vector<int>ans;
		slover.solve(ans);
		for(int i=0;i<ans.size();i++){
			int r,c,v;
			decode(ans[i],r,c,v);
			puzzle[r][c]=v+1;
		}
		for(int i=0;i<9;i++){
		for(int j=0;j<9;j++)
		printf("%d",puzzle[i][j]);
		printf("\n");
		}
	}
	return 0;
}

　　

转载于:https://www.cnblogs.com/jie-dcai/p/3934452.html