Preparation Phase
1, GitHub address
2, PSP form
| psp2.1 | Estimated time consuming (minutes) | The actual time-consuming (minutes) |
|---|---|---|
| plan | 30 | 30 |
| Estimate how much time this task requires | 20 | 15 |
| Develop | 150 | 120 |
| Needs analysis (including learning new technologies) | 240 | 300 |
| Generate design documents | 20 | 30 |
| Design Review | 10 | 10 |
| Code specifications (development of appropriate norms for the current development) | 120 | 120 |
| Specific design | 30 | 30 |
| Specific coding | 240 | 180 |
| Code Review | 20 | 15 |
| Test (self-test, modify the code, submit modifications) | 120 | 240 |
| report | 30 | 60 |
| testing report | 30 | 50 |
| Calculate quantities | 20 | 30 |
| Later summarized, and process improvement plan | 60 | 60 |
| total | 1140 | 1290 |
Initial stage
1, thinking
Baidu Encyclopedia Description:
Sudoku is a disk nine palaces, each palace is divided into nine small cell. Given certain conditions known digital and solving this eighty-one grid, using logic and reasoning, enter the number 1-9 on the other spaces. 1-9 so that each number appears only once in each row, each column and each palace, also known as "squares."
For never contacted Guo Shu separate me, this is a very, very, very difficult topic, because I could not even the most basic Sudoku rules are unclear. Even if it's only three logical judgment, but solving a Sudoku squares, still need some time.
2, thinking
First of all judgments, with three direct access to an array of bool result legitimacy.
if(row[x][i] == false && col[y][i] == false && little_shudu[area][i] == false)
So then, if we can get a result grid can be filled.
After getting the legitimacy of the box, we traversed by DFS, backtracking
shudu[x][y] = i;
row[x][i] = true;
col[y][i] = true;
if(gx != 1 || gy != 1)
little_shudu[area][i] = true;
dfs(index + 1,maxscan,jie,gx,gy);
shudu[x][y] = 0;
row[x][i] = false;
col[y][i] = false;
if(gx != 1 || gy != 1)
little_shudu[area][i] = false;3, module
1, an input module
Since the requirement is input from input.txt, the input modules are as follows:
void input(int x,int gx,int gy)
{
memset(row,false,sizeof(row));
memset(col,false,sizeof(col));
memset(little_shudu,false,sizeof(little_shudu));
for(int i = 0 ; i < x ; i++){
for(int j = 0 ; j < x ; j++){
fscanf(fp, "%d", &shudu[i][j]);
int temp_i = i / gx;
int temp_j = j / gy;
int area = temp_i * gx + temp_j;
if(shudu[i][j] != 0){
int temp = shudu[i][j];
row[i][temp] = true;
col[j][temp] = true;
if(gx != 1 || gy != 1 ){
little_shudu[area][temp] = true;
}
}
}
}
}2, the output module
Obtained only by entering the number of files and solved, and then outputs a file output.txt
void print(int x)
{
for(int i = 0 ; i < x ; i++)
{
for(int j = 0 ; j < x ; j++)
{
if(j == x - 1) {
fprintf(fp, "%d\n", shudu[i][j]);
}
else {
fprintf(fp, "%d ", shudu[i][j]);
}
}
}
fprintf(fp, "\n");
}3, solution module
That is, first check the legality of the calculation filled in.
void dfs(int index , int maxscan,int jie,int gx , int gy)
{
if(index >= maxscan)
{
print(jie);
return ;
}
int x = index / jie ;
int y = index % jie ;
int temp_i = x / gx;
int temp_j = y / gy;
int area = temp_i * gx + temp_j;
if(shudu[x][y] == 0)
{
for(int i = 1 ; i <= jie ; i++)
{
if(row[x][i] == false && col[y][i] == false && little_shudu[area][i] == false)
{
shudu[x][y] = i;
row[x][i] = true;
col[y][i] = true;
if(gx != 1 || gy != 1)
little_shudu[area][i] = true;
dfs(index + 1,maxscan,jie,gx,gy);
shudu[x][y] = 0;
row[x][i] = false;
col[y][i] = false;
if(gx != 1 || gy != 1)
little_shudu[area][i] = false;
}
}
}
else{
dfs(index + 1,maxscan,jie,gx,gy);
}
}4, performance testing
I feel like okay, Li Jue unknown.
5, run the test
Three grids done!
Four grids done!
(Intermediate 568 is omitted)
Seven grids done!
Of course, would be complete without squared!
Total code:
#include "stdafx.h"
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include<string>
#include <cstring>
#include <set>
#include <map>
#include <cmath>
#include <stack>
#include <queue>
#include <fstream>
using namespace std;
const int maxn = 1000;
const int sizen = 16;
int shudu[sizen][sizen];//存储数独
bool row[sizen][sizen];//row[i][j]判断i行j是否被填过
bool col[sizen][sizen];//col[i][j]判断i列j是否被填过
bool little_shudu[sizen][sizen];//little_sudu[i][j]判断从左往右开始的第i个小数独中，j是否被填过；
FILE* fp;
void input(int x, int gx, int gy) {
memset(row, false, sizeof(row));
memset(col, false, sizeof(col));
memset(little_shudu, false, sizeof(little_shudu));
for (int i = 0; i < x; i++) {
for (int j = 0; j < x; j++) {
scanf("%d", &shudu[i][j]);
int temp_i = i / gx;
int temp_j = j / gy;
int area = temp_i * gx + temp_j;
if (shudu[i][j] != 0) {
int temp = shudu[i][j];
row[i][temp] = true;
col[j][temp] = true;
if (gx != 1 || gy != 1) {
little_shudu[area][temp] = true;
}
}
}
}
}
void print(int x) {
for (int i = 0; i < x; i++) {
for (int j = 0; j < x; j++) {
if (j == x - 1) {
printf("%d\n", shudu[i][j]);
}
else {
printf("%d ", shudu[i][j]);
}
}
}
printf("\n");
}
void dfs(int index, int maxscan, int jie, int gx, int gy) {
if (index >= maxscan) {
print(jie);
return;
}
int x = index / jie;
int y = index % jie;
int temp_i = x / gx;
int temp_j = y / gy;
int area = temp_i * gx + temp_j;
if (shudu[x][y] == 0) {
for (int i = 1; i <= jie; i++) {
if (row[x][i] == false && col[y][i] == false && little_shudu[area][i] == false) {
shudu[x][y] = i;
row[x][i] = true;
col[y][i] = true;
if (gx != 1 || gy != 1)
little_shudu[area][i] = true;
dfs(index + 1, maxscan, jie, gx, gy);
shudu[x][y] = 0;
row[x][i] = false;
col[y][i] = false;
if (gx != 1 || gy != 1)
little_shudu[area][i] = false;
}
}
}
else {
dfs(index + 1, maxscan, jie, gx, gy);
}
}
int main(int argc, char *argv[]) {
int pan, size, jie, gx, gy;
int in, out;
for (int i = 0; i < argc; i++)
{
if (strlen(argv[i]) == 1)
{
if (i == 2)
size = atoi(argv[i]);
if (i == 4)
pan = atoi(argv[i]);
}
else if (argv[i][0] == '-' && argv[i][1] == 'i') {
i++;
in = i;
}
else if (argv[i][0] == '-' && argv[i][1] == 'o') {
i++;
out = i;
}
}
freopen(argv[in], "r", stdin);
freopen(argv[out], "w", stdout);
jie = size;
if (size == 4) {
gx = gy = 2;
}
else if (size == 6) {
gx = 2;
gy = 3;
}
else if (size == 8) {
gx = 4;
gy = 2;
}
else if (size == 9) {
gx = gy = 3;
}
else gx = gy = 1;
for (int i = 0; i < pan; i++) {
input(size, gx, gy);
dfs(0, jie*jie, jie, gx, gy);
}
return 0;
}