【日常学习】【非指针链表】Uva11988 - Broken Keyboard (a.k.a. Beiju Text)题解
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这道题目拖了好几天,因为鄙人有两大天敌——链表和树TUT看了这个题材知道原来链表可以不用指针写,不过原理也是一样的,相当于是用数组模拟了个链表而不实用结构体,结构体里的指针就换成了两个变量cur和last了。这道题目本来测出来非常奇怪和合因为UVA AC HDU TLE SPOJ RE我正在奇怪,才发现同名的不同题目有三道TUT
题目的详解已经写在了注释里,上代码:
#include<cstdio>#include<cstring>using namespace std;const int maxn=100000+5;int last,cur,next[maxn];char s[maxn];int main(){while (scanf("%s",s+1)==1){int n=strlen(s+1);//输入从s[1]开始存储 last=cur=0;next[0]=0;//0位置的下一个数初始化为0 ,会改变 for (int i=1;i<=n;i++){//按照字符的编号排列顺序输出字符,顺序是排列好的编号序列,该序列储存在next中 char ch=s[i];if (ch=='[') cur=0;else if (ch==']') cur=last;else {next[i]=next[cur];//现在字符位置的下一个是当前光标的下一个【未插入之前当前光标下一个】 a|c next[cur]=i;//当前光标的下一个是现在插入的字符位置,即此时光标还是在原来的前一个字符后面 a|bc if (cur==last) last=i;//如果当前已是最后,最后编号更新 last指向最后一个 next[last]为空 先比较再移动cur cur=i;//移动光标到当前插入字符后面 ab|c }}for (int i=next[0];i!=0;i=next[i]) printf("%c",s[i]);//注意占位符,另外其实next数组应该初始化为0 printf("\n");}return 0;}
——高祖尝徭咸阳,纵观,观始皇帝,喟然太息曰:“嗟乎,大丈夫当如此也!”
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