课上碰到的题 要交了不太能写出来能不能参考一下

Question

如题初学碰上的题 怎么写出提示 请把源代码直接甩我脸上求求

Answer 1

如果对你有帮助记得采纳一下谢谢~

参考代码如下

#include<stdio.h>
#include<stdlib.h>
#include<memory.h>
#include<math.h>
#include<time.h>
#include <windows.h>
#include<string.h>
#define MAX 999

struct 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("┃  ");
        else
            printf("┃ %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0
            
            if(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   容易\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,81);//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");
}
}

参考文章：https://blog.csdn.net/qq_45218336/article/details/107720395