1. 程式人生 > >UVA11988 Broken Keyboard (a.k.a. Beiju Text)

UVA11988 Broken Keyboard (a.k.a. Beiju Text)

看到大一練習題,睡前水一水~~~

Problem B

Broken Keyboard (a.k.a. Beiju Text)

You're typing a long text with a broken keyboard. Well it's not so badly broken. The only problem with the keyboard is that sometimes the "home" key or the "end" key gets automatically pressed (internally).

You're not aware of this issue, since you're focusing on the text and did not even turn on the monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).

In Chinese, we can call it Beiju. Your task is to find the Beiju text.

Input

There are several test cases. Each test case is a single line containing at least one and at most 100,000 letters, underscores and two special characters '[' and ']'. '[' means the "Home" key is pressed internally, and ']' means the "End" key is pressed internally. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each case, print the Beiju text on the screen.

Sample Input

This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University

Output for the Sample Input

BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University
Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University

Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!
#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#define N 100009
using namespace std;

char s[N];
int next[N];//存第i號後面是幾號~

int main()
{
    while(~scanf("%s",s+1))
    {
        next[0]=0;
        int cur=0,la=0;
        int len=strlen(s+1);

        for(int i=1;i<=len;i++)
        {
            char c=s[i];

            if(c=='[')
            cur=0;
            else if(c==']')
            cur=la;
            else
            {
                next[i]=next[cur];
                next[cur]=i;
                if(la==cur) la=i;
                cur=i;
            }
        }

        int i;
        for(i=next[0];i!=0;i=next[i])
        printf("%c",s[i]);
        cout<<endl;

    }
    return 0;
}