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:

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.

2391197625999911254265113123401234401234598346樣例輸出
NOYES
解題報(bào)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(biāo)記是否是一個(gè)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(chuàng)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(gè)單詞作為前綴,比如119,119895            return 1;        int id = str[i]-'0'; //數(shù)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(gè)單詞    return 0;}void dealTrie(Trie* T) //清理內(nèi)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(rèn)YES        char s[15];        for(int i=0;i<n;i++){            getchar();            scanf("%s",s);            if(!flag) //把數(shù)據(jù)讀完                continue;            if(createTrie(s)){                flag=0;            }        }        if(flag)            printf("YES/n");        else            printf("NO/n");        dealTrie(root); //釋放內(nèi)存，否則超內(nèi)存    }    return 0;}