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  • poj 2676 Sudoku

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    Sudoku
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 16894   Accepted: 8229   Special Judge

    Description

    Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task.

    Input

    The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

    Output

    For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

    Sample Input

    1
    103000509
    002109400
    000704000
    300502006
    060000050
    700803004
    000401000
    009205800
    804000107

    Sample Output

    143628579
    572139468
    986754231
    391542786
    468917352
    725863914
    237481695
    619275843
    854396127

    Source

     
     
    思路:

    DFS不为0的点1-9枚举,为0的跳过

    本来没什么好说的,一个教训:循环的判别条件不要出现任何计算

    15234481 njczy2010 2676 Accepted 700K 1547MS G++ 2083B 2016-03-06 09:55:21
      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 #include <algorithm>
      5 #include <stack>
      6 #include <cctype>
      7 #include <vector>
      8 #include <cmath>
      9 #include <map>
     10 #include <queue>
     11 
     12 #define ll long long
     13 
     14 using namespace std;
     15 
     16 int T;
     17 int ans[100][100];
     18 char s[100][100];
     19 int flag;
     20 
     21 struct PP
     22 {
     23     int x;
     24     int y;
     25 };
     26 
     27 int ok(int x,int y,int v)
     28 {
     29     int i,j;
     30     i = x;
     31     for(j = 1;j <= 9;j++){
     32         if(ans[i][j] == v) return 0;
     33     }
     34     j = y;
     35     for(i = 1;i <= 9;i++){
     36         if(ans[i][j] == v) return 0;
     37     }
     38     int rr = (x-1)/3*3;
     39     int cc = (y-1)/3*3;
     40     for(i = rr + 1 ;i <= rr + 3 ;i++  ){
     41         for(j =  cc +1 ;j <= cc + 3 ;j++  ){
     42             if(ans[i][j] == v) return 0;
     43         }
     44     }
     45     return 1;
     46 }
     47 
     48 void dfs(int x,int y)
     49 {
     50     int i;
     51     if(flag == 1){
     52         return;
     53     }
     54     if(x == 10){
     55         flag = 1;
     56         return;
     57     }
     58     int ff;
     59     if(ans[x][y] == 0){
     60         ff = 0;
     61         for(i = 1;i <= 9 ;i++){
     62             if(ok(x,y,i) == 1){
     63                 ans[x][y] = i;
     64                 ff = 1;
     65                 if(y == 9){
     66                     dfs(x+1,1);
     67                 }
     68                 else{
     69                     dfs(x,y+1);
     70                 }
     71                 if(flag == 1) return;
     72                 ans[x][y] = 0;
     73             }
     74         }
     75         if(ff == 0) return;
     76     }
     77     else{
     78         if(y == 9){
     79             dfs(x+1,1);
     80             if(flag == 1) return;
     81         }
     82         else{
     83             dfs(x,y+1);
     84             if(flag == 1) return;
     85         }
     86     }
     87 }
     88 
     89 int main()
     90 {
     91     //freopen("in.txt","r",stdin);
     92     scanf("%d",&T);
     93     for(int ccnt=1;ccnt<=T;ccnt++){
     94     //while(scanf("%d%s%s",&n,a,b)!=EOF){
     95         flag = 0;
     96         int i,j;
     97         for(i = 1;i <= 9;i++){
     98             scanf("%s",s[i] + 1);
     99         }
    100         for(i = 1;i <= 9;i++){
    101             for(j = 1;j <= 9;j++){
    102                 ans[i][j] = s[i][j] - '0';
    103             }
    104         }
    105         dfs(1,1);
    106         for(i = 1;i <= 9;i++){
    107             for(j = 1;j <= 9;j++){
    108                 printf("%d",ans[i][j]);
    109             }
    110             printf("
    ");
    111         }
    112     }
    113     return 0;
    114 }
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  • 原文地址:https://www.cnblogs.com/njczy2010/p/5246609.html
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