Question

我想要一个停车场管理系统的JAVA源程序

Answer 1

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 3 // 停车场最大容量
#define price 5 //每小时5元
typedef struct time {
int hour;
int min;
} Time;// 定义时间结构体
typedef struct node {
char num[10];
Time reach;
Time leave;
} CarNode;// 定义车辆信息结构体
typedef struct NODE {
CarNode * stack[MAX + 1];
int top;
} SeqStackCar;
typedef struct car {
CarNode *data;
struct car *next;
} QueueNode;
typedef struct Node {
QueueNode *head;
QueueNode *rear;
} LinkQueueCar;
void InitStack(SeqStackCar *s) {
int i;
s->top = 0;
for (i = 0; i <= MAX; i++)
s->stack[s->top] = NULL;
}// 栈的初始化
int InitQueue(LinkQueueCar *Q) {
Q->head = (QueueNode *) malloc(sizeof (QueueNode));
if (Q->head != NULL) {
Q->head->next = NULL;
Q->rear = Q->head;
return (1);
} else return (-1);
}// 队列的初始化
void PRINT(CarNode *p, int room) { // 车辆收费
int A1, A2, B1, B2;
printf("\n车辆离开的时间:");
scanf("%d:%d", &(p->leave.hour), &(p->leave.min));
printf("\n离开车辆的车牌号为:");
puts(p->num);
printf("\n其到达时间为: %d:%d", p->reach.hour, p->reach.min);
printf("\n离开时间为: %d:%d", p->leave.hour, p->leave.min);
A1 = p->reach.hour;
A2 = p->reach.min;
B1 = p->leave.hour;
B2 = p->leave.min;
printf("\n应交费用为: %2.1f元", ((B1 - A1)*60 + (B2 - A2)) * price);
free(p);
}
int Arrival(SeqStackCar *Enter, LinkQueueCar *W) {
CarNode *p;
QueueNode *t;
p = (CarNode *) malloc(sizeof (CarNode));
printf("\n请输入车牌号(例:豫B1234):");
scanf("%d",&p->num);
if (Enter->top < MAX) {
Enter->top++;
printf("\n车辆在车场第%d位置.", Enter->top);
printf("\n车辆到达时间:");
scanf("%d:%d", &(p->reach.hour), &(p->reach.min));
Enter->stack[Enter->top] = p;
return (1);
} else {
printf("\n该车须在便道等待!有车位时进入车场");
t = (QueueNode *) malloc(sizeof (QueueNode));
t->data = p;
t->next = NULL;
W->rear->next = t;
W->rear = t;
return (1);
}
}// 车辆的到达登记
void Leave(SeqStackCar *Enter, SeqStackCar *Temp, LinkQueueCar *W) { //车辆的离开
int room;
CarNode *p, *t;
QueueNode *q;

if (Enter->top > 0) // 判断车场是否为空
{
while (1) {
printf("\n请输入车在车场的位置/1--%d/：", Enter->top);
scanf("%d", &room);
if (room >= 1 && room <= Enter->top) break;
else printf("\n 输入有误,请重输: ");

}
while (Enter->top > room) // 把要删除的车辆的前面的车开出来，进临时栈。
{
Temp->top++;
Temp->stack[Temp->top] = Enter->stack[Enter->top];
Enter->stack[Enter->top] = NULL;
Enter->top--;
}
p = Enter->stack[Enter->top]; // 把要删除的车辆节点赋给p。
Enter->stack[Enter->top] = NULL;
Enter->top--;
while (Temp->top >= 1) // 再把临时栈里德车辆进停车场
{
Enter->top++;
Enter->stack[Enter->top] = Temp->stack[Temp->top];
Temp->stack[Temp->top] = NULL;
Temp->top--;
}
PRINT(p, room); // 调用计费函数计费
if ((W->head != W->rear) && Enter->top < MAX) //如果便道上有车，则再开进停车场。
{
q = W->head->next;
t = q->data;
Enter->top++;
printf("\n便道的%s号车进入车场第%d位置.", t->num, Enter->top);
printf("\n请输入%s号车进入车场的时间:", t->num);
scanf("%d:%d", &(t->reach.hour), &(t->reach.min));
W->head->next = q->next;
if (q == W->rear) W->rear = W->head;
Enter->stack[Enter->top] = t;
free(q);
} else printf("\n便道里没有车.\n");
} else printf("\n车场里没有车.");
}
void List1(SeqStackCar *S){ int i;
if (S->top > 0) {
printf("\n车场:");
printf("\n 位置 到达时间 车牌号\n");
for (i = 1; i <= S->top; i++) {
printf(" %d ", i);
printf(" %d:%d ", S->stack[i]->reach.hour, S->stack[i]->reach.min);
puts(S->stack[i]->num);
}
} else printf("\n车场里没有车");
}//显示车场里的车辆情况
void List2(LinkQueueCar *W) {
QueueNode *p;
int i;
p = W->head->next;
if (W->head != W->rear) {
printf("\n等待车辆的号码为:");
for (i = 1; (p != NULL); i++) {
printf("\n第 %d 车辆.", i);
puts(p->data->num);
p = p->next;
}
} else printf("\n便道里没有车.");

printf("\n");
}//显示便道上的车辆情况
void List(SeqStackCar S, LinkQueueCar W) {
int flag, tag;
flag = 1;
while (flag) {
printf(" 查看车辆列表显示: ");
printf("\n 1.车场列表\n 2.便道列表\n 3.返回主菜单\n");
printf("\n请选择 1~3:");
while (1) {
scanf("%d", &tag);
if (tag >= 1 && tag <= 3) break;
else printf("\n 输入有误,请重新选择 1~3:");
}
switch (tag) {
case 1:List1(&S);
break;
case 2:List2(&W);
break;
case 3:flag = 0;
system("cls");
break;
default: break;
}
}
}//显示,遍历
int main() {
SeqStackCar Enter, Temp;
LinkQueueCar Wait;
int ch;
InitStack(&Enter);
InitStack(&Temp);
InitQueue(&Wait);
while (1) {
printf("\n ############ 欢迎使用停车场系统.############\t\n\n");
printf("\n\t * 1. 车辆到达登记.\t\n");
printf("\n\t * 2. 车辆离开登记.\t\n");
printf("\n\t * 3. 车辆列表显示.\t\n");
printf("\n\t * 4. 退出系统.\t\n\n");
while (1) {
printf(" 请选择: ");
scanf("%d", &ch);
if (ch >= 1 && ch <= 4)break;
else printf("\n 输入有误，请重新选择： 1~4: ");
}
switch (ch) {
case 1:Arrival(&Enter, &Wait);
break;
case 2:Leave(&Enter, &Temp, &Wait);
break;
case 3:List(Enter, Wait);
break;
case 4:exit(0);
default: break;
}
}
}// 自定义函数

Answer 2

可以借助Baidu Hi示意我
有时间可能完成你所面临的任务
差不多的要求也可能示意我

ES:\\4A5D6082E9AF04F9C868A53D1F1BF6F3

Answer 3

这是c，不是java

Answer 4

这个有没有电路图啊，求助。一个小时能告诉我不，谢谢了