| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 20113 | Accepted: 9628 | 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
#include<iostream> #include<cstdio> #include<cstring> #include<sstream> #include<algorithm> #include<queue> #include<vector> #include<cmath> #include<map> #include<stack> #include<set> #include<fstream> #include<memory> #include<list> #include<string> using namespace std; typedef long long LL; typedef unsigned long long ULL; #define MAXN 10 #define INF 1000000009 /* 数独游戏，9*9的方格 要求 每行，每一列，还有九个3*3的方格（如图） 中都出现1—9的全部数字 DFS回溯法搜索 难点在于如何保证九个 3 3的方格中出现1到9的全部数字 对每个方格找到对应的区域然后检查 */ struct node { int x, y; node(int _x,int _y):x(_x),y(_y){} }; vector<node> v; char g[MAXN][MAXN],ans[MAXN][MAXN]; bool judge(int x, int y,char c) { int tx = x / 3 * 3, ty = y / 3 * 3; for (int i = 0; i < 9; i++) { if (g[i][y] == c) return false; if (g[x][i] == c) return false; if (g[tx + i/3][ty + i-i/3*3] == c) return false; } return true; } bool dfs(int x) { if (x == v.size()) return true; for (int i = 1; i <= 9; i++) { if (!judge(v[x].x, v[x].y, i+'0')) continue; //cout << v[x].x << ' ' << v[x].y << ' ' << char(i + '0') << endl; g[v[x].x][v[x].y] = i + '0'; if (dfs(x + 1)) return true; } g[v[x].x][v[x].y] = '0'; return false; } int main() { int T; scanf("%d", &T); while (T--) { v.clear(); for (int i = 0; i < 9; i++) { scanf("%s", g[i]); for (int j = 0; j < 9; j++) { if(g[i][j]=='0') v.push_back(node(i, j)); } } dfs(0); for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { printf("%c", g[i][j]); } printf(" "); } } return 0; }
改进版 ：空间换时间
把judge改成inline能节省300ms
| 680K | 860MS |
#include<iostream> #include<cstdio> #include<cstring> #include<sstream> #include<algorithm> #include<queue> #include<vector> #include<cmath> #include<map> #include<stack> #include<set> #include<fstream> #include<memory> #include<list> #include<string> using namespace std; typedef long long LL; typedef unsigned long long ULL; #define MAXN 10 #define INF 1000000009 /* 数独游戏，9*9的方格 要求 每行，每一列，还有九个3*3的方格（如图） 中都出现1—9的全部数字 DFS回溯法搜索 难点在于如何保证九个 3 3的方格中出现1到9的全部数字 对每个方格找到对应的区域然后检查 */ struct node { int x, y; node(int _x,int _y):x(_x),y(_y){} }; vector<node> v; char g[MAXN][MAXN],ans[MAXN][MAXN]; bool col[MAXN][MAXN], row[MAXN][MAXN], grid[MAXN][MAXN][MAXN]; bool judge(int x, int y,int i) { int tx = x / 3 * 3, ty = y / 3 * 3; if (!row[x][i] && !col[y][i] && !grid[tx][ty][i]) return true; return false; } bool dfs(int x) { if (x == v.size()) return true; for (int i = 1; i <= 9; i++) { if (!judge(v[x].x, v[x].y, i)) continue; col[v[x].y][i] = row[v[x].x][i] = grid[v[x].x / 3 * 3][v[x].y / 3 * 3][i] = true; g[v[x].x][v[x].y] = i + '0'; if (dfs(x + 1)) return true; col[v[x].y][i] = row[v[x].x][i] = grid[v[x].x / 3 * 3][v[x].y / 3 * 3][i] = false; } g[v[x].x][v[x].y] = '0'; return false; } int main() { int T; scanf("%d", &T); while (T--) { v.clear(); memset(grid, false, sizeof(grid)); memset(row, false, sizeof(row)); memset(col, false, sizeof(col)); for (int i = 0; i < 9; i++) { scanf("%s", g[i]); for (int j = 0; j < 9; j++) { if(g[i][j]=='0') v.push_back(node(i, j)); else { int tx = i / 3 * 3, ty = j / 3 * 3; // cout << tx <<' '<< ty << endl; row[i][g[i][j] - '0'] = col[j][g[i][j] - '0'] = grid[tx][ty][g[i][j] - '0'] = true; } } } dfs(0); for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { printf("%c", g[i][j]); } printf(" "); } } return 0; }