poj3630/hdu1671 Phone List

                                                                     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:

• Emergency 911
• Alice 97 625 999
• 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      

想法：暑假回爐重造第一天，無限TLE and MLE，菜到不行。簡單的字典樹模闆題，一開始以為是hdu的題，動态指針建樹後忘記銷毀指針一直MLE；後面改了一直TLE，才發現是poj的3630，也是醉了。要用數組來建樹才行。不會這玩意隻能百度去學了，真的是菜。

CODE:

hdu1671版（指針建樹）

#include<stdio.h>
#include<string.h>
struct trie{
        trie *next[10];
        int v;
        trie(){
                v=1;
                for(int i=0;i<10;i++) next[i]=NULL;
        }
}*root;
void create_trie(char *str){
        int len=strlen(str);
        trie *p=root,*q;
        for(int i=0;i<len;i++){
                int id=str[i]-'0';
                if(p->next[id]==NULL){
                        q=new trie;
                        p->next[id]=q;
                        p=p->next[id];
                }
                else{
                        p->next[id]->v++;
                        p=p->next[id];
                }
        }
        p->v=-1;
}
int find_trie(char *s){
        int len=strlen(s);
        trie *q=root;
        for(int i=0;i<len;i++){
                int id=s[i]-'0';
                q=q->next[id];
                if(q==NULL) return 0;
                if(q->v==-1) return -1;
        }
        return -1;
}
void delete_trie(trie* q){
        if(q==NULL) return ;
        for(int i=0;i<10;i++) delete_trie(q->next[i]);
        delete q;
}
int main()
{
        int t,n,i;
        char s[15];
        scanf("%d",&t);
        while(t--){
                scanf("%d",&n);
                root=new trie;
                int flag=0;
                for(i=0;i<n;i++){
                        scanf("%s",s);
                        if(flag) continue;
                        if(find_trie(s)==-1) flag=1;
                        if(flag) continue;
                        create_trie(s);
                }
                if(flag) puts("NO");
                else puts("YES");
                delete_trie(root);
        }
}
           

poj3630（數組建樹）

#include<cstring>
#include<cstdio>
using namespace std;
const int N=1e4+5;
char a[N][11];
struct node
{
    bool flag;
    node *next[11];
    node(){
        flag=0;
        memset(next,0,sizeof(next));
    }
}trie[100000];
int num;
void Insert(node *root,char *s)
{
    int i=0;
    node *p=root;
    while(s[i]){
        int k=s[i++]-'0';
        if(p->next[k]==NULL) p->next[k]=&trie[num++];
        p=p->next[k];
    }
    p->flag=1;
}
bool Search(node *root,char *s)
{
    int i=0;
    node *p=root;
    while(s[i]){
        int k=s[i++]-'0';
        p=p->next[k];
        if(p->flag&&s[i]) return 1;
    }
    return 0;
}
int main()
{
    int t,n,i;
    scanf("%d",&t);
    while(t--){
        memset(trie,0,sizeof(trie));
        node *root=&trie[0];
        num=1;
        scanf("%d",&n);
        for(i=0;i<n;i++){
            scanf("%s",a[i]);
            Insert(root,a[i]);
        }
        int flag=0;
        for(i=0;i<n;i++){
            if(Search(root,a[i])){
                flag=1;
                break;
            }
        }
        if(!flag) puts("YES");
        else puts("NO");
    }
    return 0;
}