Sudoku
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 21612 | Accepted: 10274 | 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
题意：把一个9行9列的网格，再细分为9个3*3的子网格，要求每行、每列、每个子网格内都只能使用一次1~9中的一个数字，即每行、每列、每个子网格内都不允许出现相同的数字，填完数独。
分析：直接搜索，标记行、列、块，值得一提的是倒着搜比正着搜效率高出许多，这也算是一个技巧。
代码：
1 ////#include "bits/stdc++.h"
2 #include "cstdio"
3 #include "map"
4 #include "set"
5 #include "cmath"
6 #include "queue"
7 #include "vector"
8 #include "string"
9 #include "cstring"
10 #include "time.h"
11 #include "iostream"
12 #include "stdlib.h"
13 #include "algorithm"
14 #define db double
15 #define ll long long
16 #define vec vector<ll>
17 #define Mt vector<vec>
18 #define ci(x) scanf("%d",&x)
19 #define cd(x) scanf("%lf",&x)
20 #define cl(x) scanf("%lld",&x)
21 #define pi(x) printf("%d\n",x)
22 #define pd(x) printf("%f\n",x)
23 #define pl(x) printf("%lld\n",x)
24 #define rep(i, x, y) for(int i=x;i<=y;i++)
25 const int N = 1e6 + 5;
26 const int mod = 1e9 + 7;
27 const int MOD = mod - 1;
28 const db eps = 1e-18;
29 const db PI = acos(-1.0);
30 using namespace std;
31 int t;
32
33 int R()
34 {
35 int x=0,f=1;char ch=getchar();
36 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
37 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
38 return x*f;
39 }
40 char s[10][10];
41 int a[10][10];
42 bool ok(int ans,int x,int y)
43 {
44 for(int i=0;i<9;i++)
45 if(a[i][y]==ans) return 0;
46 for(int i=0;i<9;i++)
47 if(a[x][i]==ans) return 0;
48 int xx=x-x%3,yy=y-y%3;
49 for(int i=xx;i<xx+3;i++)
50 for(int j=yy;j<yy+3;j++)
51 if(a[i][j]==ans) return 0;
52 return 1;
53 }
54 bool okk=0;
55 void dfs(int x,int y,int cnt)
56 {
57 if(cnt==81){
58 okk=1;//满足条件后立刻结束，并标记
59 return;
60 }
61 while(a[x][y]){
62 if(y==8) x++,y=0;
63 else y++;
64 if(x==9) {okk=1;return;}//满足条件后立刻结束，并标记
65 }
66 for(int i=1;i<=9;i++){
67 if(ok(i,x,y)){
68 a[x][y]=i;
69 if(y==8) dfs(x+1,0,cnt+1);
70 else dfs(x,y+1,cnt+1);
71 if(okk) return;//满足条件后立刻结束
72 a[x][y]=0;
73 }
74 }
75 return;
76 }
77 int main()
78 {
79 t=R();
80 while(t--)
81 {
82 int cnt=81;
83 memset(a,0, sizeof(a));
84 memset(s,0, sizeof(s));
85 okk=0;
86 for(int i=0;i<9;i++)
87 {
88 scanf("%s",s[i]);
89 for(int j=0;j<9;j++){
90 a[i][j]=s[i][j]-‘0‘;
91 if(!a[i][j]) cnt--;
92 }
93 }
94 dfs(0,0,cnt);//输出即为满足条件的，结束后的情况。
95 for(int i=0;i<9;i++){
96 for(int j=0;j<9;j++){
97 printf("%d",a[i][j]);
98 }
99 puts("");
100 }
101 }
102 }