c语言数独程序缺少函数

Question

1、删除readFromFile函数的函数体，考察学生的文件读写能力；
2、删除Solve函数；
3、删除solveSudoku函数的函数体。其中：
a)参数const char a[][COL]表示初盘二维数组；
参数char b[][COL]表示解的二维数组
需要将其添加进去

/*
 *       数独学生端程序
 *           左妍 
 *    计算机科学与技术系 
 *         2017/6/12 
 */
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#define ROW 9
#define COL 9
#define CORRECT 0
#define WRONG -1 
#define MAXNUM 1000 /*数独游戏个数*/  
int checkSudoku(const char a[][COL]){
    int i,j,x,y;
    for( i = 0; i < ROW ; i++){
        for(j = 0 ; j < COL ; j++){        
            for(x = 0 ; x < COL ; x++)
                if(a[i][x] == a[i][j] && x != j ) 
                    return WRONG;
            
            for(x = 0 ; x < ROW ; x++)
                if(a[x][j] == a[i][j] && x != i ) 
                    return WRONG;
        
            for(x = i/3*3 ; x < i/3*3+3 ; x ++)
                for(y = j/3*3 ; y < j/3*3+3 ; y++)
                     if(a[x][y] == a[i][j] && (x != i && y != j)) 
                         return WRONG;         
        }    
    }
    return CORRECT;
}
void printSudoku(const char a[][COL]){
    int i,j;
    printf("\n ┌────┬────┬────┐\n"); 
    for( i = 0 ; i < ROW ; i++){
        if(i != 0 && i%3 == 0) printf(" ├────┼────┼────┤\n");
        printf(" │");
        for( j = 0 ; j < COL ; j++){
            if(j != 0 && j%3 == 0) printf("  │");
            if(a[i][j])
                printf(" %d",a[i][j]);
            else printf("  ");
        }
        printf("  │\n");
    }
    printf(" └────┴────┴────┘\n");
}
void readFromFile(char a[][COL], const char filename[]){
    // To Be Solved
    // 从文件filename中读取到的数独初盘存储到数组a[][COL]中
 
}
void solveSudoku(const char a[][COL], char b[][COL]){
    // To Be Solved 
    // 参数const char a[][COL]表示初盘二维数组；
    // 参数char b[][COL]表示解的二维数组。 
}
int main(){
    char starting_grid[ROW][COL]={0};
    char result[ROW][COL]={0};
    char filename[50];
    int rr;
    double time_from, time_to, time_sum = 0;
    int i;
    for( i = 0; i < MAXNUM ; i++){
        sprintf(filename,"sudoku\\%04d.txt",i);
        readFromFile(starting_grid,filename);
        printf("\n\n             *%4d  *",i);
        printSudoku(starting_grid);
        time_from = clock();
        solveSudoku(starting_grid,result);    
        time_to = clock();
        time_sum += time_to - time_from;    
        printSudoku(result);
        rr = checkSudoku(result);
        if( rr == WRONG){
            printf("Something goes wrong...\n");
            return WRONG;    
        } 
        else printf("            Correct!!!\n");
        printf("  ==============================");
    }
    printf("\n Congretulations! ToTal Time:%fs\n",time_sum / CLOCKS_PER_SEC);
    return 0;
}

Answer 1

基于深度搜索实现版本

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<time.h>
#define ROW 9
#define COL 9
#define CORRECT 0
#define WRONG -1 
#define MAXNUM 1000 /*数独游戏个数*/  
#define bool int
#define true 1
#define false 0
bool DFS(char b[][COL], int n);
int checkSudoku(const char a[][COL]){
    int i,j,x,y;
    for( i = 0; i < ROW ; i++){
        for(j = 0 ; j < COL ; j++){    

            for(x = 0 ; x < COL ; x++) // 列
                if(a[i][x] == a[i][j] && x != j ) 
                    return WRONG;

            for(x = 0 ; x < ROW ; x++) // 行
                if(a[x][j] == a[i][j] && x != i ) 
                    return WRONG;

            for(x = i/3*3 ; x < i/3*3+3 ; x ++) // 9宫格内
                for(y = j/3*3 ; y < j/3*3+3 ; y++)
                     if(a[x][y] == a[i][j] && (x != i && y != j)) 
                         return WRONG;     

        }    
    }
    return CORRECT;
}
void printSudoku(const char a[][COL]){
    int i,j;
    printf("\n ┌───────┬───────┬───────┐\n"); 
    for( i = 0 ; i < ROW ; i++){
        if(i != 0 && i%3 == 0) printf(" ├───────┼───────┼───────┤\n");
        printf(" │");
        for( j = 0 ; j < COL ; j++){
            if(j != 0 && j%3 == 0) printf(" │");
            if(a[i][j]!=' ')
                printf(" %d",(a[i][j]-'0')); /*怀疑这里出问题,原来是 printf(" %d",a[i][j]);*/
            else 
                printf(" %c",a[i][j]);
        }
        printf(" │\n");
    }
    printf(" └───────┴───────┴───────┘\n");
}
void readFromFile(char a[][COL], const char filename[]){
    // To Be Solved
    // 从文件filename中读取到的数独初盘存储到数组a[][COL]中
    FILE *fp = NULL;
     if((fp = fopen(filename, "r"))==NULL){
        printf("Fail to open file!\n");
        return;
    }
    int i=0, j=0;
    while(!feof(fp)){
        fscanf(fp,"%s",a[i]);
        for(j=0;i<ROW&&j<COL;j++)
            if(a[i][j]=='0')
                a[i][j]=' ';
        i++;
    }
    fclose(fp);
}
void writeToFile(char a[][COL], const char filename[]){
    // To Be Solved
    // 数独初盘存储到数组a[][COL]中存储到文件filename
    FILE *fp;
    if((fp = fopen(filename, "w+") )==NULL){   /*打开文件写模式*/
       printf("cannot open the file.\n");   /*判断文件是否正常打开*/
       return;
    }
    for(int i=0;i<ROW;i++){
        for(int j=0;j<COL;j++)
            if(a[i][j] == ' ')
                fputc('0',fp);
            else
                fputc(a[i][j],fp);
        fputc('\n',fp);
    }
    fclose(fp);
}
void solveSudoku(const char a[][COL], char b[][COL]){
    // To Be Solved 
    // 参数const char a[][COL]表示初盘二维数组；
    // 参数char b[][COL]表示解的二维数组。 
    for(int i=0;i<ROW;i++)
        for(int j=0;j<COL;j++)
            b[i][j] = a[i][j];
    DFS(b,0);
}
int main(){
    char starting_grid[ROW][COL]={0};
    char result[ROW][COL]={0};
   
    char filename[50];
    int rr;
    double time_from, time_to, time_sum = 0;
    int i;  
    for( i = 0; i < MAXNUM ; i++){
        sprintf(filename,"E:\\C\\Test\\sudoku\\%04d.txt",i);
        readFromFile(starting_grid,filename);
        printf("\n\n         *%4d  *",i);
        printSudoku(starting_grid);

        time_from = clock();
        solveSudoku(starting_grid,result);    
        time_to = clock();

        time_sum += time_to - time_from;    

        printSudoku(result);
        rr = checkSudoku(result);

        if(rr == WRONG){
            printf("Something goes wrong...\n");
            //return WRONG;  
            continue;  
        } 
        else 
            printf("            Correct!!!\n");
        printf("  ==============================");
    }
    printf("\n Congretulations! ToTal Time:%fs\n",time_sum / CLOCKS_PER_SEC);
    system("pause");
}
bool Check(char b[][COL],int n, char key){
    int x,y,i,j;
    x = n / 9;
    y = n % 9;
    // 检查纵坐标方向
    for(i=0;i<ROW;i++)
        if(b[i][y] == key && x!=i) 
            return false;
    // 检查横坐标方向
    for(j=0;j<COL;j++)
         if(b[x][j] == key && y!=j) 
            return false;
    //  检查小宫格
    for(i = x/3*3 ; i < x/3*3+3 ; i ++) // 9宫格内
        for(j = y/3*3 ; j < y/3*3+3 ; j++)
                if(b[i][j]==key && (x != i && y != j)) 
                    return false;
    return true;     
}
bool DFS(char b[][COL], int n){
    int x,y;
    x = n / 9;
    y = n % 9;
    if(n>80) 
        return true;// 退出条件
    if(b[x][y]!=' '){ //不为空
       return DFS(b,n+1);
    }else{
        char c;
        //遍历各个情况:枚举
        for(int i=1;i<=9;i++){
            c = i+'0';
            if(Check(b, n, c)){ 
                b[x][y] = c;
                // 继续搜索
                bool res = DFS(b,n+1);
                if(res==true) 
                    return res;
                /* 如果构造不成功，还原当前位 */
                b[x][y] = ' ';
            }
        }
    }
    return false;
}

我做了一些改动，辅助输出好看点