UVa 11988 Broken Keyboard(鏈表的應用)

標籤：acm   劉汝佳   uva   資料結構   鏈表   

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__textHappy_Birthday_to_Tsinghua_University

題意  電腦鍵盤的home鍵和end鍵壞了  會在你不注意時自動按下

給你一個輸入序列 ‘[‘代表home鍵   ‘]‘代表end鍵  要求輸出螢幕上對應的輸出

用鏈表儲存每個位置的字元c和下一個位置的編號next  最後一個字元的next為0

並用cur表示游標的移動

#include<cstdio>#include<cstring>using namespace std;const int N = 100005;char s[N], c;int cur, last, l;struct Node{ char c; int next;}  lis[N];int main(){    while (~scanf ("%s", s + 1))    {        l = strlen (s + 1);        lis[0].next = last = cur = 0;        for (int i = 1; i <= l; ++i)        {            c = s[i];            if (c == '[') cur = 0;            else if (c == ']') cur = last;            else            {                lis[i].c = c;                lis[i].next = lis[cur].next;                lis[cur].next = i;                if (cur == last) last = i;                cur = i;            }        }        for (int i = lis[0].next; i != 0; i = lis[i].next)            printf ("%c", lis[i].c);        printf ("\n");    }    return 0;}