一、功能：
（1）输出排名
从文件中读出数据，保存在单链表中，并利用选择排序对数据进行排序后输出。
（2）开始游戏
进入该游戏系统后，系统先随机创建一个唯一解的数独，并解出该数独答案。
（3）选择难度
根据玩家选择的难度，挖空系统已经解出的数独，比如选择简单（给出40个数），则把系统已经解出的答案随机挖空41格，挖空处用数字0代替。
（4）给出数独并开始作答
玩家得到题目后进行解答（解答需将数独完整写一遍，以空格分割每列，以回车分割每行），最后系统根据答案判断玩家解答是否正确，正确则输出所用时间，如果该玩家是新纪录，则还要记录到文件，如果玩家解答错误，则失败并给出正确答案。
功能结构图：
二、涉及数据结构：
十字双向循环链表
三、涉及算法：
DLX算法
如何将数独转化为精准覆盖问题，并用DLX算法去解
数据结构体设计：
①玩家游戏记录信息
②存储数独结点信息
四、实验环境：
使用语言：C
使用软件：Microsoft Visual C++ 6.0
五、功能截图：
六、具体代码：
#include<stdio.h>
#include<stdlib.h>
#include<memory.h>
#include<math.h>
#include<time.h>
#include <windows.h>
#include<string.h>
#define MAX 999struct Node{
Node *up,*down,*left,*right,*colPtr,*rowPtr;//指向row,col对应的行对象和列对象
int row_num;//记录行数，行专属(从1开始)
int col_elemCount;//记录该列元素个数，列专属
};struct player{//玩家信息结点
int m ;
int s ;
char name[20];
int level ;
player* next;
};int row_size=593;//行数
int col_size = 324;//列数
int result[81];//存放结果行的栈
int index = 0;//栈指针
int sudoku[81] = {0};//存放数独
int time_start = 0;
int time_end = 0;//起始时间void init(Node* head){head->left = head;head->right = head;head->up = head;head->down = head;for(int k = 0;k < row_size;k++){//创建行对象(头插)Node* newNode = (Node*)malloc(sizeof(Node));newNode->up = head;newNode->down = head->down;newNode->left = newNode;newNode->right = newNode;newNode->down->up = newNode;head->down = newNode;newNode->row_num = row_size-k;newNode->col_elemCount = 0;//借用，作为标志}
}
void init_col(Node* head){/************初始化行列对象*******************/for(int j = 0;j < col_size;j++){//创建列对象(头插)Node* newNode = (Node*)malloc(sizeof(Node));newNode->right = head->right;newNode->left = head;newNode->down = newNode;newNode->up = newNode;newNode->right->left = newNode;head->right = newNode;newNode->col_elemCount = 0;//列元素个数初始为0}
}void link(Node* head,int** matrix){
/***********插入结点*************/
Node *current_row, *current_col, *current;//当前行对象,当前列对象,当前节点
current_row = head;
for(int row = 0;row<row_size;row++){current_row = current_row->down;current_col = head;for(int col = 0;col<col_size;col++){current_col = current_col->right;if(matrix[row][col] == 0)continue;/*****插入结点，结点的行和列都用尾插法来与行列对象链接*****/Node* newNode = (Node*)malloc(sizeof(Node));newNode->colPtr = current_col;//设置节点对应的行和列newNode->rowPtr = current_row;/**********列尾插************/newNode->down = current_col;newNode->up = current_col->up;newNode->up->down = newNode;current_col->up = newNode;//链接当前节点到列双向链尾端if(current_row->col_elemCount == 0){//行的这个为0,说明该行还没有元素，行双向链表不把行对象包含进来(便于后面覆盖)current_row->right = newNode;newNode->left = newNode;newNode->right = newNode;current_row->col_elemCount++;//此为标志，为了不把行对象包含进来}current = current_row->right;newNode->left = current->left;/**********行尾插************/newNode->right = current;newNode->left->right = newNode;current->left = newNode;//链接当前节点到行双向链尾端current_col->col_elemCount++;//该列元素加1}}
}int** create_matrix()//将数独转换为01矩阵
{int** matrix = (int**)malloc(row_size*sizeof(int*));//申请二维数组空间for(int m=0;m<row_size;m++)matrix[m] = (int*)malloc(col_size*sizeof(int));for(int r=0;r<row_size;r++)for(int c=0;c<col_size;c++)matrix[r][c] = 0;//初始化int i = 0;for (int x = 0; x < 81; x++){int val = sudoku[x];if (val != 0){//有值则插入一行matrix[i][x] = 1;matrix[i][81 + x/9*9 + val -1] = 1;matrix[i][162 + x%9*9 + val -1] = 1;matrix[i][243 + (x/9/3*3+x%9/3)*9 +val -1] = 1;i++;continue;}for (int j = 0; j < 9; j++){//没值则插入9行（1~9）matrix[i][x] = 1;matrix[i][81 + x/9*9 + j] = 1;matrix[i][162 + x%9*9 +j] = 1;matrix[i][243 + (x/9/3*3+x%9/3)*9 +j] = 1;i++;}}return matrix;
}void cover(Node* cRoot){//覆盖cRoot->left->right = cRoot->right;cRoot->right->left = cRoot->left;//删除该列对象，及该列结点的每行结点Node *i, *j;i = cRoot->down;while (i != cRoot){j = i->right;while (j != i){j->down->up = j->up;j->up->down = j->down;j->colPtr->col_elemCount--;j = j->right;}i = i->down;}
}void recover(Node* cRoot){//回溯
Node *i, *j;i = cRoot->up;while (i != cRoot){j = i->left;while (j != i){j->colPtr->col_elemCount++;j->down->up = j;j->up->down = j;j = j->left;}i = i->up;}cRoot->right->left = cRoot;cRoot->left->right = cRoot;}bool search(Node* head){if(head->right == head){return true;}Node *cRoot, *c;int minSize = MAX;//最少列元素个数for(c = head->right; c != head; c = c->right)//先选择列元素最少的列对象（提高效率）{if (c->col_elemCount < minSize){minSize = c->col_elemCount;cRoot = c;if (minSize == 1)//1是最小break;if (minSize == 0)//有一列为空,失败return false;}}cover(cRoot);Node *current_row,*current;for (current_row = cRoot->down; current_row != cRoot; current_row = current_row->down){result[index]=current_row->rowPtr->row_num;//将该行加入result中(行数)index++;for (current = current_row->right; current != current_row; current = current->right){cover(current->colPtr);}if (search(head))//递归return true;for (current = current_row->left; current != current_row; current = current->left)recover(current->colPtr);index--;//发现该行不符合要求,出栈}recover(cRoot);return false;
}int* to_sudoku(int** matrix){//01矩阵转化为数独int* done = (int*)malloc(81*sizeof(int));int temp[162]={0};for(int i = 0;i<81;i++){for(int j = 0;j<162;j++)temp[j] = matrix[result[i]-1][j];int pos = 0,val = 0;for(int k = 0;k<81;k++){if(temp[k] == 1)break;pos++;}for(int m = 81;m<162;m++){if(temp[m] == 1)break;val++;}done[pos]=val%9+1;}return done;
}void print(int* answer){//打印数独
printf("┏━━┳━━┳━━┳━━┳━━┳━━┳━━┳━━┳━━┓\n");for(int i = 0;i<81;i++){if(answer[i]==0)printf("┃ ");elseprintf("┃ %d",answer[i]);if(i==80){printf("┃ ");printf("\n");printf("┗━━┻━━┻━━┻━━┻━━┻━━┻━━┻━━┻━━┛\n");}else if((i+1)%9==0){printf("┃ ");printf("\n");printf("┣━━╋━━╋━━╋━━╋━━╋━━╋━━╋━━╋━━┫\n");}}
}bool isRight(int* flag,int x,int y,int val){//判断填入的数是否合法for(int i = 0;i<324;i++)if(flag[(x-1)*9+(y-1)] == 0 &&flag[81+(x-1)*9 + val -1] == 0 &&flag[162+(y-1)*9 + val -1] == 0 &&flag[243+((x-1)/3*3 + (y-1)/3+1)*9 +val -1] == 0 )return true;
return false;
}void create_sudoku(){for(int j = 0;j<81;j++)sudoku[j] = 0;//初始化int flag[324]={0};int x,y,val;//行号、列号、值int i=0;while(i<17){//一个数独中有17个数及以上时有唯一解x = rand()%9+1;y = rand()%9+1;val = rand()%9+1;if(isRight(flag,x,y,val)){i++;sudoku[(x-1)*9+(y-1)]=val;flag[(x-1)*9+(y-1)] = 1;flag[81+(x-1)*9 + val -1] = 1;flag[162+(y-1)*9 + val -1] = 1;flag[243+((x-1)/3*3 + (y-1)/3+1)*9 +val -1] = 1;}}
}void sudoku_level(int* answer,int count){//难度int x,y;//行号、列号int num=0;srand(time(NULL));for(int i = 0;i<81;i++)sudoku[i] = answer[i];while(num<(81-count)){//挖空x = rand()%9+1;y = rand()%9+1;if(sudoku[(x-1)*9+(y-1)] != 0){sudoku[(x-1)*9+(y-1)] = 0;num++;}}
}bool receiver(int* player_res){//接收玩家答案for(int i=0;i<81;i++){scanf("%d",&player_res[i]);if(!(player_res[i]>=1&&player_res[i]<=9)){//0 表示玩家放弃fflush(stdin);return false;}}return true;
}int get_time(){//获得当前时间秒time_t t;t=time(NULL);return t;
}void ready(){print(sudoku);printf("你有5秒钟观察时间\n");for(int i = 0;i<5;i++){printf("● ");Sleep(1000);}printf("\n");printf("观察结束，计时开始，请开始作答。（输入除1~9外，视为放弃作答）\n");printf("==========================================================================\n");
}bool judge(int* player_res,int* answer){//判断玩家答案for(int i = 0;i<81;i++)if(player_res[i] != answer[i])return false;
return true;
}void record(player info){//记录FILE* fp;int M = MAX,S = MAX,LEVEL = MAX;char NAME[20];char remove[100] = {" "};//用于记录长度固定化，方便更新记录//通过这种方法，可以直接在一个文件中更新数据，不必要全篇读—改—写，直接修改一行int c = 0;if((fp = fopen("record.txt","r+"))==NULL){//文件在cpp同目录下printf("文件不存在，保存失败！");//虽然会自动生成文件，but以防万一return;}setbuf(fp,NULL);//设置缓冲区rewind(fp);c = ftell(fp);//记录当前行的开头指针位置while(fscanf(fp,"%s %d:%d %d",NAME,&M,&S,&LEVEL) != EOF){if(!strcmp(NAME,info.name) && LEVEL == info.level){//strcmp比较相同返回0if(info.m<M || (info.m==M&&info.s<S)){//如果是新纪录，则更新fseek(fp,c,SEEK_SET);fputs(remove,fp);//覆盖旧记录fseek(fp,c,SEEK_SET);//回到该记录的开头位置fprintf(fp,"%s %d:%d %d",info.name,info.m,info.s,info.level);//写入文件fflush(fp);//清除缓冲区return;}return;//不是新纪录就不插入}fscanf(fp,"\n");//读取换行c = ftell(fp);}fputs(remove,fp);//先覆盖固定长度的区域fseek(fp,c,SEEK_SET);//回到覆盖的区域首部fprintf(fp,"%s %d:%d %d",info.name,info.m,info.s,info.level);//在覆盖的区域内插入记录fseek(fp,0,SEEK_END);//指向尾部fprintf(fp,"\n");//插入换行符fclose(fp);
}player* get_record(int level){//返回玩家记录的单向链表头结点FILE* fp;int M = MAX,S = MAX,LEVEL = MAX;char NAME[20];player* head = (player*)malloc(sizeof(player));head->next = NULL;if((fp = fopen("record.txt","r"))==NULL){printf("文件不存在！");system("pause");exit(1);}setbuf(fp,NULL);//设置缓冲区rewind(fp);while(fscanf(fp,"%s %d:%d %d",NAME,&M,&S,&LEVEL) != EOF){if(LEVEL == level){player* p = (player*)malloc(sizeof(player));//采用链表strcpy(p->name,NAME);p->m = M;p->s = S;p->next = head ->next;head->next = p;} }
fclose(fp);
return head;
}void order(player* head){//单链表排序
player* p;
player* q;
int temp1;
int temp2;
char temp3[20];
for(p = head->next;p != NULL;p = p->next)for(q = p->next;q != NULL;q = q->next)if(p->m>q->m||(p->m==q->m&&p->s>q->s)){//对换两个结点的内容temp1 = p->m;temp2 = p->s;strcpy(temp3,p->name);p->m = q->m;p->s = q->s;strcpy(p->name,q->name);q->m = temp1;q->s = temp2;strcpy(q->name,temp3);}}void show(player* easy,player* normal,player* hard){//输出排行int no=1;player* p1 = easy->next;player* p2 = normal->next;player* p3 = hard->next;
printf("==========================================================================\n");printf("\t\t 简单\t\t\t 一般\t\t\t 困难\n");while(p1 != NULL || p2 != NULL || p3 != NULL){printf("NO.%d",no++);if(p1 != NULL){printf("\t\t%s\t%d:%d\t",p1->name,p1->m,p1->s);p1 = p1->next;}if(p2 != NULL){printf("\t%s\t%d:%d\t",p2->name,p2->m,p2->s);p2 = p2->next;}if(p3 != NULL){printf("\t%s\t%d:%d\t",p3->name,p3->m,p3->s);p3 = p3->next;}printf("\n");}
}void re_init(Node* head){//重新初始化行和列对象
Node* p;
for(p = head->down;p!=head;p=p->down)p->col_elemCount = 0;if(head->right == head){//如果有解，则列对象会在search中被删除，需要重新链接列对象init_col(head);return;}for(p = head->right;p!=head;p=p->right){//无解时,则恢复列对象p->col_elemCount = 0;p->down = p;p->up = p;}
}void main(){
int player_res[81]={0};
int choice;
int** matrix;//存放数独的01矩阵
int* answer;//存放答案
player* easy;//容易难度排行
player* normal;//简单难度排行
player* hard;//困难难度排行
player info ;//玩家信息Node* head =(Node*)malloc(sizeof(Node));
init(head);init_col(head);//初始化行列对象，建立十字双向循环链表
srand(time(NULL));
bool flag = false;//有无解
while(true){while(!flag){flag=true;create_sudoku();//创建17个数的数独（17个数及以上唯一解）matrix = create_matrix();//得到该数独的01矩阵link(head,matrix);//将01矩阵转化为十字双向循环链表（转化为精准覆盖问题）if(!search(head))//得到答案行（result）flag = false;//如果不幸无解，则重新构建数独（经本人测试，无解的概率大概为1/1000）index = 0;//初始化result栈re_init(head);//重新初始化十字链表}flag = false;
answer = to_sudoku(matrix);//得到答案printf(" 数独\n");
printf("==========================================================================\n\n");
printf("1.开始游戏\n");
printf("2.查看排名\n");
printf("3.退出\n\n");
printf("==========================================================================\n");
printf("请选择:");
scanf("%d",&choice);
switch(choice){
case 1:int option;printf("玩家名:");scanf("%s",&info.name);if(strlen(info.name)>20){printf("名字太长！\n");break;}printf("请选择游戏难度: 1.简单\t2.一般\t3.困难\n");scanf("%d",&option);printf("\n");switch(option){case 1:sudoku_level(answer,40);//挖空答案ready();time_start = get_time();if(!receiver(player_res)){printf("\n您已放弃作答!\t正确答案为:\n\n");print(answer);break;}time_end = get_time();info.m = (time_end-time_start)/60;info.s = (time_end-time_start)%60;info.level = 1;if(judge(player_res,answer)){printf("回答正确!\t用时: %d:%d\n",info.m,info.s);record(info);}else {printf("\n回答错误!\t正确答案为:\n\n");fflush(stdin);print(answer);}break;case 2:sudoku_level(answer,35);time_start = get_time();ready();if(!receiver(player_res)){printf("\n您已放弃作答!\t正确答案为:\n\n");print(answer);break;}time_end = get_time();info.m = (time_end-time_start)/60;info.s = (time_end-time_start)%60;info.level = 2;if(judge(player_res,answer)){printf("回答正确!\t用时: %d:%d\n",info.m,info.s);record(info);}else {printf("回答错误!\t正确答案为:\n\n");fflush(stdin);print(answer);}break;case 3:sudoku_level(answer,30);time_start = get_time();ready();if(!receiver(player_res)){printf("\n您已放弃作答!\t正确答案为:\n\n");print(answer);break;}time_end = get_time();info.m = (time_end-time_start)/60;info.s = (time_end-time_start)%60;info.level = 3;if(judge(player_res,answer)){printf("回答正确!\t用时: %d:%d\n",info.m,info.s);record(info);}else {printf("回答错误!\t正确答案为:\n\n");fflush(stdin);print(answer);}break;default:printf("no option!\n");fflush(stdin);break;}break;
case 2:easy = get_record(1);normal = get_record(2);hard = get_record(3);order(easy);order(normal);order(hard);show(easy,normal,hard);break;
case 3:printf("\n拜拜~\n\n");exit(0);
default:printf("no option!\n");fflush(stdin);break;
}
system("pause");
system("cls");
}
}
