UVa 11988 Broken Keyboard (a.k.a. Beiju Text) (指针的"思想")

11988 - Broken Keyboard (a.k.a. Beiju Text)

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=3139

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

思路：

首先，这道题可以用链表实现（用deque更快）

但我们不妨用一种"指针"的思想来解决问题。

开一个next[]数组用以标记下一个预输出的位置，从而隐性地完成对原字符串的重排。

最后，通过遍历next[]数组完成输出。

完整代码：

/*0.059s*/

#include<cstdio>

char s[100005];
int next[100005];///next[i]指向了下一个要输出的字符的位置

int main()
{
	while (gets(s))
	{
		int now = 0, front = 0, back = 0;
		next[0] = 0;
		for (int i = 0; s[i]; ++i)
			if (s[i] == '[') now = 0, front = next[0];
			else if (s[i] == ']') now = back, front = 0;
			else
			{
				next[now] = i + 1; ///记录now位置后面的要输出的位置
				now = i + 1;
				if (front) next[i + 1] = front; /// 模拟：把该字符加到队首
				else next[i + 1] = 0, back = i + 1; ///0为end标记
			}
		for (int i = next[0]; i; i = next[i])
			putchar(s[i - 1]);
		putchar(10);
	}
	return 0;
}