hdu 1022 Train Problem I（栈的操作）；

Train Problem I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39281    Accepted Submission(s): 14788

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 3213 123 312

 

Sample Output

Yes.inininoutoutoutFINISHNo.FINISH

能否根据栈的操作将第一个序列转换成第二个序列；如132  321   应输出"Yes.";火车进出栈顺序如下：一号火车进栈，三号火车进栈，三号火车出栈，二号火车进栈， 二号火车出栈，一号火车出栈；开个flag数组记录火车进栈还是出栈，值为1进栈，0出栈；

感觉数据也够水的，开个10的flag也过了......;

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#include <stack>#define MAX 8000+5using namespace std;int main(){    int n;    string a, b;    while(cin >> n >> a >> b){        stack<char> s;        int flag[1000];               //亲测开10也能过，不过为了保险起见还是开的大一点；        int k=0, j=0;        memset(flag, -1, sizeof(flag));        for(int i=0; i<n; i++){            s.push(a[i]);            flag[++k]=1;            while(!s.empty() && s.top()==b[j]){                flag[++k]=0;                j++;                s.pop();            }        }        if(s.empty()){            cout << "Yes.\n";            for(int i=1; i<=k; i++)                if(flag[i]) cout << "in\n";                else cout << "out\n";                cout << "FINISH\n";        }        else{            cout << "No.\n" << "FINISH\n";        }    }    return 0;}