HUD 1022-Train Problem I

HUD 1022-Train Problem I

原题：

As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can’t leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.

输入要求：

The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.

输出要求：

The output contains a string “No.” if you can’t exchange O2 to O1, or you should output a line contains “Yes.”, and then output your way in exchanging the order(you should output “in” for a train getting into the railway, and “out” for a train getting out of the railway). Print a line contains “FINISH” after each test case. More details in the Sample Output.

输入样例：

3 123 321

3 123 312

输出样例：

Yes.

in

in

in

out

out

out

FINISH

No.

FINISH

翻译：

随着新学期的到来，如今Ignatius火车站很繁忙，许多学生想要坐火车回学校（因为Ignatius火车站的火车是全世界最快的）。但是这里有个问题，在所有的火车站点上只有一个铁轨，所以，所有火车都是从一边进入，再从另一边出去。对于这个问题，如果火车A第一个进入铁轨，然后火车B在火车A前离开，一直到火车B离开，火车A都不能离开。这个图片展示了这个问题。现在在火车站上有最多有9个火车，所有的火车有一个编号（从1到n），火车以O1次序进入铁轨，你的任务是确定火车是否能以O2的次序离开。

输入包含多个测试数据。每组测试数据是由一个整数（列车的数量）和两个字符串，火车进入的次序是：O1，火车离开的次序：O2，到文件尾结束。

如果你不能交换O1和O2，你要输出一行“No”，否则你应该输出一行“Yes”，然后输出你交换次序的方式（火车入轨为”in”，火车出轨为”out”），最后打印一行“FINISH”。

用c语言写：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct Node{
int num;
struct Node *next;
};

typedef struct Node *NODE;
typedef NODE Stack;
Stack CreatStack();
void MakeEmpty(Stack s);
void push(int n, Stack s);
void pop(Stack s);
int top(Stack s);
int isEmpty(Stack s);

int main(){
int n;
int a,b,c;
//a指向下一次要进入的火车
//b指向下一次要出去的火车
char in[15] = { 0 };
char out[15] = { 0 };
int step[25];
//数组step和c用来记录火车的进入和出去，默认-1，进入1，出去0
while (scanf("%d %s %s", &n,in,out)!=EOF){
memset(step,-1,sizeof(step));
Stack s = CreatStack();
a = 0;
b = 0;
c = 0;
while (a < n){
//如果栈不是空，或者栈顶不符合当前需要出去的火车吗，就进入
if (isEmpty(s) || top(s) != out[b]-'0'){
push(in[a] - '0', s);
step[c++] = 1;
a++;
}
//循环判断可以出去的
while (!isEmpty(s) && top(s) == out[b]-'0'){
pop(s);
step[c++]=0;
b++;
}
}
if (a == n ){
if (isEmpty(s)){
printf("Yes.\n");
int i = 0;
for(i=0;i<c;i++){
if(step[i]==1){
printf("in\n");
}
else{
printf("out\n");
}
}
}
else{
printf("No.\n");
}
}
printf("FINISH\n");
}
return 0;
}

//检查栈是否为空
int isEmpty(Stack s){
return s->next == NULL;
}
//清空栈
void MakeEmpty(Stack s){
while (!isEmpty(s)){
pop(s);
}
}
//建立栈
Stack CreatStack(){
Stack s;
s = malloc(sizeof(struct Node));
s->next = NULL;
MakeEmpty(s);
return s;
}
//进栈
void push(int n, Stack s){
NODE new_node;
new_node = malloc(sizeof(struct Node));
new_node->num = n;
new_node->next = s->next;
s->next = new_node;
}
//出栈
void pop(Stack s){
NODE temp_node = s->next;
s->next = temp_node->next;
free(temp_node);
}
//返回栈顶
int top(Stack s){
if (!isEmpty(s)){
return s->next->num;
}
return 0;
}