杭电1022——Train Problem I（栈的应用）

Problem Description
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.

Input
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.

Output
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.

Sample Input
3 123 321
3 123 312

Sample Output
Yes.
in
in
in
out
out
out
FINISH
No.
FINISH

Hint
Hint

For the first Sample Input, we let train 1 get in, then train 2 and train 3.
So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.
In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.
Now we can let train 3 leave.
But after that we can’t let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.
So we output “No.”.

主要算法：
扫描字符串2。如果当前字符与栈顶元素相等，则出栈，并且扫描字符串2中下一个字符；如果栈为空或者字符串2中扫描的当前字符与栈顶元素不相等，则字符串1中字符按顺序压入栈中，直到满足字符串2当前扫描字符和栈顶字符相等。再次，栈顶元素出栈，并且扫描字符串2中下一个字符。如此重复下去，如果最终栈为空，则输出Yes.如果栈顶元素和字符串2中当前扫描的字符不等，且字符串1中的元素已经全部进栈，则输出No.
注意：每个case完了之后要清空栈。

#include<stdio.h>#include<string.h>#define MAX 50typedef struct{    char data[MAX];    int top;}Stack;void Initial(Stack *s);char getTop(Stack *s);int Push(Stack *s,char ch);int Pop(Stack *s);int main(){    int n;    char s1[MAX],s2[MAX];    char in_out[MAX][5];    Stack s;    Initial(&s);    while(scanf("%d%s%s",&n,s1,s2)!=EOF)    {        int i=0,j=0,k=0,flag=1;        memset(in_out,0,MAX*5*sizeof(char));        Push(&s,s1[i++]);        strcpy(in_out[k++],"in");        while(i<n||j<n)        {            while(s.top==0||s2[j]!=getTop(&s))            {                if(i<n)                {                    Push(&s,s1[i++]);                    strcpy(in_out[k++],"in");                }                else                {                    flag=0;                    break;                }            }            if(flag==0)                break;            Pop(&s);            strcpy(in_out[k++],"out");            j++;        }        if(s.top==0)        {            printf("Yes.\n");            for(i=0;i<k;i++)                printf("%s\n",in_out[i]);            printf("FINISH\n");        }        else        {            printf("No.\nFINISH\n");        }        s.top=0;//注意要将栈清空    }    return 0;}void Initial(Stack *s){    s->top=0;}char getTop(Stack *s){    return s->data[s->top-1];}int Push(Stack *s,char ch){    if(s->top>=MAX)        return 0;    else    {        s->data[s->top++]=ch;        return 1;    }}int Pop(Stack *s){    if(s->top>=MAX)        return 0;    else    {        s->data[--s->top]=0;        return 1;    }}