UVA11988 Broken Keyboard (a.k.a. Beiju Text)【输入输出】

最新推荐文章于 2025-12-19 16:34:09 发布

转载最新推荐文章于 2025-12-19 16:34:09 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/tigerisland/p/7563617.html

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#数据结构与算法 #c/c++

本文解析了UVA11988 Broken Keyboard (Beiju Text)问题，介绍了如何处理一个特殊键盘输入的问题，该键盘偶尔会自动触发Home或End键。通过构造一个特殊的数组nexti[]来模拟光标的移动，从而得到正确的屏幕显示文本。

You’re typing a long text with a broken keyboard. Well it’s not so badly broken. The only problemwith 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 themonitor! 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 100,000letters, underscores and two special characters ‘[’ and ‘]’. ‘[’ means the “Home” key is pressedinternally, and ‘]’ means the “End” key is pressed internally. The input is terminated by end-of-file(EOF).

Output

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

Sample Input

This_is_a_[Beiju]_text

[[]][][]Happy_Birthday_to_Tsinghua_University

Sample Output

BeijuThis_is_a__text

Happy_Birthday_to_Tsinghua_University

问题链接：UVA11988 Broken Keyboard (a.k.a. Beiju Text)。

问题简述：参见上文。

问题分析：破损的键盘，字符串中的字符"["相当于"Home"键（光标跳到输入行的开始），字符"]"相当于"End"键（光标跳到输入行的结束）。

程序说明：

使用数组nexti[]，nexti[i]=n表示s[i]右边的字符为s[n]。

程序的关键是设置数组nexti[]各个元素值，该数组如同一个链表一般。

题记：目前的计算机内存地址是一维线性编码的，所以如果能用数组来表示问题，计算速度相对较快。

AC的C++语言程序如下：

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

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 100000 + 2;
char s[N];
int nexti[N];

int main()
{
    int current, last;

    while(fgets(s+1, N, stdin) != NULL) {
        current = last = 0;
        nexti[0] = 0;
        int i = 1;
        while(s[i] != '\n') {
            if(s[i] == '[')
                current = 0;
            else if(s[i]==']')
                current = last;
            else {
                nexti[i] = nexti[current];
                nexti[current] = i;
                if(last == current)
                    last = i;
                current = i;
            }
            i++;
       }

        for(i = nexti[0]; i; i = nexti[i])
            printf("%c", s[i]);
        putchar('\n');
    }

    return 0;
}

转载于:https://www.cnblogs.com/tigerisland/p/7563617.html