首页 > 代码库 > HDU 1671 Phone List(Trie的应用与内存释放)
HDU 1671 Phone List(Trie的应用与内存释放)
Phone List
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18217 Accepted Submission(s): 6120
Problem Description
Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let’s say the phone catalogue listed these numbers:
1. Emergency 911
2. Alice 97 625 999
3. Bob 91 12 54 26
In this case, it’s not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob’s phone number. So this list would not be consistent.
1. Emergency 911
2. Alice 97 625 999
3. Bob 91 12 54 26
In this case, it’s not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob’s phone number. So this list would not be consistent.
Input
The first line of input gives a single integer, 1 <= t <= 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 <= n <= 10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.
Output
For each test case, output “YES” if the list is consistent, or “NO” otherwise.
Sample Input
2
3
911
97625999
91125426
5
113
12340
123440
12345
98346
Sample Output
NO
YES
题目链接:HDU 1671
嗯还是一道简单的Trie,但是对于内存的要求比较高= =不管是malloc或new出来的,用完一定要释放不然……MLE数次。说是Trie其实是学习一下如何进行有效率地释放……
注意题目中可能给的单词顺序不同会出现两种存在前缀的情况
例1、输入911 后输入911000,这个比较简单,逐一插入911000的时候若节点存在则检查是否此时的节点的flag为true即这个节点是一个结尾点。
例2、输入911000 后输入911,办法有很多,我是用一个any来记录后面的911插入时是否出现过未出现的字母,显然在节点不存在new的时候any就要变成1了。
代码:
#include <stdio.h>#include <bits/stdc++.h>using namespace std;const int N=10;const int M=15;struct Trie{ Trie *nxt[N]; bool flag; Trie() { for (int i=0; i<N; ++i) nxt[i]=NULL; flag=false; }};Trie *L;bool update(char s[]){ int i,len=strlen(s); int indx; bool is_end=false; bool any=false;//至少存在一个没出现过的单词 Trie *cur=L; for (i=0; i<len; ++i) { indx=s[i]-‘0‘; if(!cur->nxt[indx]) { Trie *one=new Trie(); any=true; cur->nxt[indx]=one; cur=one; } else { cur=cur->nxt[indx]; if(cur->flag)//过程中遇到结尾单词,前缀存在 is_end=true; } } if(!any) is_end=true; cur->flag=true; return is_end;}void deleTrie(Trie *L){ Trie *cur=L; for (int i=0; i<N; ++i) if(cur->nxt[i]) deleTrie(cur->nxt[i]); delete cur;}char s[M];int main(void){ int tcase,n; scanf("%d",&tcase); while (tcase--) { bool flag=true; L=new Trie(); scanf("%d",&n); while (n--) { scanf("%s",s); if(flag) { if(update(s)) flag=false; } } puts(flag?"YES":"NO"); deleTrie(L); } return 0;}
HDU 1671 Phone List(Trie的应用与内存释放)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。