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:
Emergency 911Alice 97 625 999Bob 91 12 54 26In 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.
輸入The first line of input gives a single integer, 1 ≤ t ≤ 10, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 ≤ n ≤ 100000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.輸出For each test case, output "YES" if the list is consistent, or "NO" otherwise.樣例輸入2391197625999911254265113123401234401234598346樣例輸出NOYES解題報告：用字典樹
code：
#include<iostream>#include<stdio.h>#include<queue>#include<vector>#include<stack>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int maxn=100005;const int MAX=10;typedef struct node{ struct node *next[MAX]; int flag; //標記是否是一個單詞}Trie;Trie *root;/*root要初始化root=(Trie *)malloc(sizeof(Trie));root->flag=0;for(int i=0;i<MAX;i++){ root->next[i]=NULL;}*/int createTrie(char *str) //創建一棵字典樹,與查找合并{ int len = strlen(str); Trie *p = root, *q; for(int i=0; i<len; i++) { if(p->flag==1) //查找1；說明已有一個單詞作為前綴,比如119,119895 return 1; int id = str[i]-'0'; //數字字符 if(p->next[id] == NULL) { q = (Trie *)malloc(sizeof(Trie)); q->flag = 0; //初始v==1 for(int j=0; j<MAX; j++) q->next[j] = NULL; p->next[id] = q; } p = p->next[id]; } for(int i=0;i<MAX;i++){ //查找2；判斷該單詞是否是其它單詞的前綴，如119895，119 if(p->next[i]!=NULL) return 1; } p->flag=1; //一個單詞 return 0;}void dealTrie(Trie* T) //清理內存root{ for(int i=0;i<MAX;i++) { if(T->next[i]!=NULL) dealTrie(T->next[i]); } free(T);}int main(){ // freopen("input.txt","r",stdin); int t,n; scanf("%d",&t); while(t--){ root=(Trie *)malloc(sizeof(Trie)); //初始化 root->flag=0; for(int i=0;i<MAX;i++){ root->next[i]=NULL; } scanf("%d",&n); int flag=1; //默認YES char s[15]; for(int i=0;i<n;i++){ getchar(); scanf("%s",s); if(!flag) //把數據讀完 continue; if(createTrie(s)){ flag=0; } } if(flag) printf("YES/n"); else printf("NO/n"); dealTrie(root); //釋放內存，否則超內存 } return 0;}
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