2676 Sudoku

虽然数据通过了，但是超时

这是一定的，那么多循环，发现一个规律，用数组可以省时间，而且少错误

我的代码——超时

#include "iostream"
#include "string.h"
#include "algorithm"
using namespace std;
int map[10][10];
int dir[9][2]={{1,1},{1,4},{1,7},{4,1},{4,4},{4,7},{7,1},{7,4},{7,7}};


bool chang(){
   int set[10],set1[10],set2[10],i,j;
   memset(set,0,sizeof(set));
   memset(set1,0,sizeof(set1));
   memset(set2,0,sizeof(set2));
   for(i=1;i<=9;i++){
     int a=dir[i-1][0],b=dir[i-1][1];
     for(j=1;j<=9;j++){
      int ansa=a+(j-1)/3,ansb=b+(j-1)%3;
      if(map[i][j]==0)return 0;
      if(set1[map[i][j]]){return 0;}
      if(set[map[j][i]]){return 0;}
      if(set2[map[ansa][ansb]]){return 0;}
      set2[map[ansa][ansb]]=1;
      set1[map[i][j]]=1;
      set[map[j][i]]=1;
     }
    memset(set,0,sizeof(set));
    memset(set1,0,sizeof(set1));
    memset(set2,0,sizeof(set2));
   }
   return 1;
}

bool check(int a,int b,int c){
  int i,tema,temb;
  tema=(a-1)/3*3+1;
  temb=(b-1)/3*3+1;
  for(i=1;i<=9;i++){
    int ansa=tema+(i-1)/3,ansb=temb+(i-1)%3;
    if(map[a][i]==c){return 0;}
    if(map[i][b]==c){return 0;}
    if(map[ansa][ansb]==c){return 0;}
  }
  return 1;
}


bool dfs(int tem){
  int a=(tem-1)/9+1,b=tem%9,j;
  if(b==0)b=9;
  if(chang()&&tem==81)return 1;
  if(map[a][b]==0){
    for(j=1;j<=9;j++){
      if(check(a,b,j)){
        map[a][b]=j;
        if(dfs(tem+1)){return 1;}
        map[a][b]=0;
      }
    }
  }
  else{
    if(dfs(tem+1)){return 1;}
  }
}

int main(){
  int ncase;
  char list[10];
  cin>>ncase;
  while(ncase--){
    int i,j;
    for(i=1;i<=9;i++){
      cin>>list;
      for(j=0;j<9;j++){
        map[i][j+1]=list[j]-'0';
      }
    }
    if(dfs(1))
     for(i=1;i<=9;i++){
     for(j=1;j<=9;j++){
       cout<<map[i][j];
     }cout<<endl;
   }
  }
}

人家的代码

#include <stdio.h>
#include <string.h>
#include <time.h>
 int rowok[9][10],colok[9][10],miyaok[9][10];　　//用来记录行列宫有哪些数字
 int sudoku[9][9],place[81][2],p,n;　　//place用来记录哪些位置需要填数
int search(int x, int y){
    int i,j;
    if(p == n){
        for(i = 0; i < 9; i++){
            for(j = 0; j < 9; j++)
                printf("%d",sudoku[i][j]);
            puts("");
        }
        return 1;
    }
    for(i = 1; i < 10; i++)
        if(!rowok[x][i] && !colok[y][i] && !miyaok[x/3*3+y/3][i]){
            sudoku[x][y] = i;
            rowok[x][i] = colok[y][i] = miyaok[x/3*3+y/3][i] = 1;
            p++;
            if(search(place[p][0],place[p][1]))
                return 1;
            p--;
            sudoku[x][y] = 0;
            rowok[x][i] = colok[y][i] = miyaok[x/3*3+y/3][i] = 0;
        }
    return 0;
}
int main (void){
    long t1;
    int i,j,T;
//    freopen("poj2676.in","r",stdin);
    scanf("%d",&T);
    while(T--){
        p = 0;
        memset(rowok,0,sizeof(rowok));
        memset(colok,0,sizeof(colok));
        memset(miyaok,0,sizeof(miyaok));
        for(i = 0; i < 9; i++){
            for(j = 0; j < 9; j++){
                scanf("%1d",&sudoku[i][j]);
                if(sudoku[i][j]){
                    rowok[i][sudoku[i][j]] = 1;
                    colok[j][sudoku[i][j]] = 1;
                    miyaok[i/3*3+j/3][sudoku[i][j]] = 1;
                }
                else{
                    place[p][0] = i;
                    place[p][1] = j;
                    p++;
                }
            }
        }
        n = p;
        p = 0;
        search(place[p][0],place[p][1]);
    }
    return 0;
}