UVa 11988 - Broken Keyboard (a.k.a. Beiju Text) 题解
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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
题目大意:你在输入文章的时候,键盘上的Home键和End键出了问题,会不定时的按下。
给你一段按键的文本,其中'['表示Home键,']'表示End键,输出这段悲剧的文本。
思路:使用链表来模拟,遇到Home键,就将后边的文本插入到这段文本的最前边,遇到
End键,就插入到这段文本的最后边。但是用链表会用到指针,过程比较繁琐。这里用一个
Next数组模拟指向,Next[i]表示当前显示屏中s[i]右边的字符下标。再用一个cur表示当前
光标的位置,last表示最后一个字符的记录位置,这样遇到End键,就能直接找到光标下一
个指向的字符位置了。
具体参考:算法竞赛入门经典(第二版)P143~144
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;char s[101000];int Next[101000];int main(){ int cur,last;//cur为光标位置,last为显示屏最后一个字符 while(~scanf("%s",s+1)) { memset(Next,0,sizeof(Next)); int len = strlen(s+1); Next[0] = 0; cur = last = 0; for(int i = 1; i <= len; i++) { if(s[i] == '[') cur = 0; else if(s[i] == ']') cur = last; else { //模拟插入链表过程 Next[i] = Next[cur];//第i个字符指向光标位置 Next[cur] = i;//光标指向下一个字符 if(cur == last)//只有光标在当前最后一个字符位置或是遇到]后才执行 last = i; cur = i;//移动光标 } } for(int i = Next[0]; i != 0; i = Next[i]) printf("%c",s[i]); printf("\n"); memset(s,0,sizeof(s)); } return 0;}
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