HDU

 HDU - 1671   Phone List

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.

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

2391197625999911254265113123401234401234598346

Sample Output

NOYES

题意：

求是否存在某串是另一个串的前缀

思路：

刚学字典树，可以用字典树，先存下所有字符串，然后按字典序从大到小排序。当添加节点时判断此时进去的字符串最后一个节点是否在另一个串中存在（此前已排序），若 cnt 值不为一则此串是某一个串的前缀。

代码:

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;//重点是简便的清空字典树的应用，用数组结构的清空效率高 struct trie{trie *Next[15];char ch;int cnt;}bee[1000005];char s[25];int f[15];int o;//记录数组个数 trie *creat(trie *root,char ch){root = &bee[o++];//赋值操作 ,相当于//root = (trie*)malloc(sizeof(root)); for(int i = 0; i < 15; i++){root->Next[i] = NULL;}root->ch = ch;root->cnt = 1;return root;}int add(trie *root,char p[]){int len=strlen(p);int k;for(int i = 0; i < len; i++){if(root->Next[p[i]-'0'] == NULL){trie *r = creat(r,p[i]);root -> Next[p[i]-'0'] = r;root = root->Next[p[i]-'0'];}else{root = root->Next[p[i]-'0'];root ->cnt++;}}return root->cnt;}struct data{char s[25];int len;}p[10005];bool cmp(data a,data b){return a.len>b.len;}int main(){int tt,t;scanf("%d",&tt);while(tt--){o = 0;memset(f,0,sizeof(f));int flag=1;scanf("%d",&t);//清空字典树 for(int i = 0; i < t*15; i++){for(int j = 0; j < 15; j++){bee[i].Next[j] = NULL;}}trie *root;root = &bee[o++];for(int i = 0; i < t; i++){//存入结构体 scanf("%s",p[i].s);p[i].len=strlen(p[i].s);}sort(p,p+t,cmp);//排序方便查找 for(int i=0;i<t;i++){int k=add(root,p[i].s);if(k>1){flag=0;break;}}if(flag)printf("YES\n");else printf("NO\n");}return 0;}