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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8575 | Accepted: 3074 |
Description
In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,
| . | 2 | 7 | 3 | 8 | . | . | 1 | . |
| . | 1 | . | . | . | 6 | 7 | 3 | 5 |
| . | . | . | . | . | . | . | 2 | 9 |
| 3 | . | 5 | 6 | 9 | 2 | . | 8 | . |
| . | . | . | . | . | . | . | . | . |
| . | 6 | . | 1 | 7 | 4 | 5 | . | 3 |
| 6 | 4 | . | . | . | . | . | . | . |
| 9 | 5 | 1 | 8 | . | . | . | 7 | . |
| . | 8 | . | . | 6 | 5 | 3 | 4 | . |
Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.
Input
The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.
Output
For each test case, print a line representing the completed Sudoku puzzle.
Sample Input
.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534. ......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3. end
Sample Output
527389416819426735436751829375692184194538267268174593643217958951843672782965341 416837529982465371735129468571298643293746185864351297647913852359682714128574936
Source
/*************************************************************************
> File Name: sudoku.cpp
> Author: ALex
> Mail: 405045132@qq.com
> Created Time: 2015年01月06日 星期二 14时46分52秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int n = 350;
const int m = 800;
const int inf = 0x3f3f3f3f;
const int N = 1000 * 1000;
bool mat[1000][1000];
char su[100];
int cnt[m];
int most;
bool ans[N];
int head;
struct Node
{
int up, down, left, right;
int col, row;
}node[N];
void init (int m)
{
memset (ans, 0, sizeof(ans));
for (int i = 0; i <= m; ++i)
{
node[i].left = i - 1;
node[i].right = i + 1;
node[i].col = i;
node[i].up = i;
node[i].down = i;
cnt[i] = 0;
}
node[0].left = m;
node[m].right = 0;
}
void remove(int c)
{
node[node[c].left].right = node[c].right;
node[node[c].right].left = node[c].left;
for (int i = node[c].down; i != c; i = node[i].down)
{
for (int j = node[i].right; j != i; j = node[j].right)
{
--cnt[node[j].col];
node[node[j].up].down = node[j].down;
node[node[j].down].up = node[j].up;
}
}
}
void resume(int c)
{
node[node[c].left].right = c;
node[node[c].right].left = c;
for (int i = node[c].up; i != c; i = node[i].up)
{
for (int j = node[i].left; j != i; j = node[j].left)
{
++cnt[node[j].col];
node[node[j].down].up = j;
node[node[j].up].down = j;
}
}
}
bool Dacing_Link_X()
{
if (node[324].right == 324)
{
return true;
}
int mins = inf;
int c;
for (int i = node[324].right; i != 324; i = node[i].right)
{
if (cnt[i] < mins)
{
mins = cnt[i];
c = i;
}
}
remove(c);
for (int i = node[c].down; i != c; i = node[i].down)
{
for (int j = node[i].right; j != i; j = node[j].right)
{
remove(node[j].col);
}
ans[node[i].row] = 1;
if (Dacing_Link_X())
{
return true;
}
for (int j = node[i].left; j != i; j = node[j].left)
{
resume(node[j].col);
}
ans[node[i].row] = 0;
}
resume(c);
return false;
}
int in_grid(int i, int j)
{
++i;
++j;
if (1 <= i && i <= 3 && 1 <= j && j <= 3)
{
return 1;
}
if (1 <= i && i <= 3 && 4 <= j && j <= 6)
{
return 2;
}
if (1 <= i && i <= 3 && 7 <= j && j <= 9)
{
return 3;
}
if (4 <= i && i <= 6 && 1 <= j && j <= 3)
{
return 4;
}
if (4 <= i && i <= 6 && 4 <=j && j <= 6)
{
return 5;
}
if (4 <= i && i <= 6 && 7 <= j && j <= 9)
{
return 6;
}
if (7 <= i && i <= 9 && 1 <= j && j <= 3)
{
return 7;
}
if (7 <= i && i <= 9 && 4 <= j && j <= 6)
{
return 8;
}
if (7 <= i && i <= 9 && 7 <= j && j <= 9)
{
return 9;
}
}
void add(int i, int j, int k)
{
int x = (i * 9 + j) * 9 + k;
mat[x][i * 9 + j] = 1;
mat[x][81 + i * 9 + k] = 1;
mat[x][162 + j * 9 + k] = 1;
int g = in_grid(i, j);
mat[x][243 + (g - 1) * 9 + k] = 1;
}
int main()
{
// printf("please enter your sudoku\n");
//
while (~scanf("%s", su))
{
if (!strcmp(su, "end"))
{
break;
}
most = 1000;
memset (mat, 0, sizeof(mat));
init(324);
int cur = 325;
for (int i = 0; i < 9; ++i)
{
for (int j = 0; j < 9; ++j)
{
if (su[i * 9 + j] == '.')
{
for (int k = 0; k < 9; ++k)
{
add(i, j, k);
}
}
else
{
int k = su[i * 9 + j] - '1';
add(i, j, k);
}
}
}
for (int i = 0; i < 729; ++i)
{
int s = cur;
int pre = cur;
for (int j = 0; j < 324; ++j)
{
if (!mat[i][j])
{
continue;
}
int pos = j;
node[cur].up = node[pos].up;
node[node[pos].up].down = cur;
node[cur].down = pos;
node[pos].up = cur;
node[cur].row = i;
node[cur].col = pos;
node[cur].left = pre;
node[pre].right = cur;
node[cur].right = s;
cnt[pos]++;
node[s].left = cur;
pre = cur;
cur++;
}
}
Dacing_Link_X();
for (int i = 0; i < 81; ++i)
{
for (int j = 0; j < 9; ++j)
{
if (ans[i * 9 + j])
{
printf("%d", j + 1);
break;
}
}
}
printf("\n");
}
return 0;
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原文地址:http://blog.csdn.net/guard_mine/article/details/42464189