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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8909 | Accepted: 3196 |
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
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node{
int x,y,cnt;
}p[100];
bool cmp(const node& a,const node& b){
return a.cnt<b.cnt;
}
int g[9][9];
int row[9][10];
int col[9][10];
int cell[9][10];
int getId(int x,int y){
return (x/3)*3 + y/3;
}
void printans(){
for(int i=0;i<9;i++){
for(int j=0;j<9;j++){
printf("%d",g[i][j]);
}
// printf("\n");
}
printf("\n");
}
int flag = 1;
int xd[90],yd[90];
void dfs(int n,int m){
if(n==m){
flag = 0;
printans();
return;
}
int small = 10;
int x,y;
int idx;
for(int v =n ;v<m;v++){
int i = p[v].x, j = p[v].y;
p[v].cnt = 0;
for(int k=1;k<=9;k++){
p[v].cnt += (row[i][k]+col[j][k]+cell[getId(i,j)][k]==0);
}
if(p[v].cnt==0)return;
if(p[v].cnt<small){
small = p[v].cnt;
idx = v;
if(small==1)break;
}
}
if(small==0||small==10) return;
swap(p[n],p[idx]);
x = p[n].x;
y = p[n].y;
for(int k=1;k<=9;k++){
if(row[x][k]+col[y][k]+cell[getId(x,y)][k]==0&&flag){
g[x][y] = k;
row[x][k]=col[y][k]=cell[getId(x,y)][k]=1;
dfs(n+1,m);
g[x][y] = 0;
row[x][k]=col[y][k]=cell[getId(x,y)][k]=0;
}
}
}
int main(){
char s[100];
while(1){
scanf("%s",s);
if(s[0]==‘e‘)break;
memset(row,0,sizeof(row));
memset(col,0,sizeof(col));
memset(cell,0,sizeof(cell));
int ct_zero = 0;
for(int u=0;u<81;u++)
{
int i=u/9;
int j=u%9;
if(s[u]>‘0‘&&s[u]<=‘9‘)g[i][j] = s[u]-‘0‘;
else {
g[i][j] = 0;
p[ct_zero].x = i;
p[ct_zero].y = j;
ct_zero++;
}
row[i][g[i][j]]=1;
col[j][g[i][j]]=1;
cell[getId(i,j)][g[i][j]]=1;
}
for(int v = 0;v<ct_zero;v++){
int i = p[v].x, j = p[v].y;
p[v].cnt = 0;
for(int k=1;k<=9;k++){
p[v].cnt += (row[i][k]+col[j][k]+cell[getId(i,j)][k]==0);
}
}
sort(p,p+ct_zero,cmp);
flag = 1;
dfs(0,ct_zero);
}
return 0;
}
如果按照每个点的选择自由度从低到高的顺序搜索会更快
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原文地址：http://www.cnblogs.com/zhjou/p/4760815.html
