Phone List

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 33442	Accepted: 9661

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:

• 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

Source

題意是打電話 如果在播出一個電話以前會打出去 就輸出no else  yes

就是一個字串如果他的前面一部分 是另一個字串 就輸出 NO  else  YEs

剛開始思路就是sort 一下長度和字串順序  因為是排序過的  所以從前往後找 找到就輸出 遍歷到最後還未找到就 輸出YES

暴力ac程式碼

#include <iostream>      //暴力sort程式碼  ac
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <map>
#include <stdlib.h>
#include <string.h>
#include <sstream>
#include <math.h>
using namespace std;
string s[10005];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--){
		int n,flag=1;
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			cin>>s[i];
		}
		sort(s+1,s+1+n);
		for(int i=1;i<n;i++){
			if(s[i+1].find(s[i])!=string::npos){
				printf("NO\n");
				flag=0;
				i=n;
			}
		}
		if(flag)printf("YES\n");
	} 
}

隨後發現正解是字典樹，用字典樹寫了一遍 ，但超時了，聽dalao講的需要用靜態連結串列 =-= 我用的動態連結串列超時很絕望。

未ac字典樹程式碼(動態連結串列)

#include <iostream>     //字典樹 動態連結串列 未ac =-=
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <map>
#include <stdlib.h>
#include <string.h>
#include <sstream>
#include <math.h>
#include <queue>
using namespace std;
int digit=0,leg;
string a[10005],temp;
struct len{
	bool flag;
	int num;
	struct len *next[10];
	len(){
		flag=false;
		num=-1;
		memset(next,0,sizeof(next));
	}
}root;
void hanshu(string s){
	int fuck=s.length();
	leg=min(leg,fuck);
	struct len *p=&root;
	for(int i=0;i<fuck;i++){
		if(p->next[s[i]-'0']==NULL){
			p->next[s[i]-'0']=new struct len;
		}
		p=p->next[s[i]-'0'];
	}
	if(p->flag==0){
		p->flag=1;
		p->num=digit++;
	}
}
int cha(string s){
	int fuck=s.length();
	struct len *p=&root;
	for(int i=0;i<fuck;i++){
		if(p->next[s[i]-'0']==0){
			return -1;
		}
		p=p->next[s[i]-'0'];
	}
	return p->num;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
    	int dee=0;
    	int n;
    	struct len cmp;
    	root=cmp;
    	digit=0;
    	leg=1000000;
    	scanf("%d",&n);
    	for(int i=1;i<=n;i++){
    		cin>>a[i];
    		hanshu(a[i]);
		}
		for(int i=1;i<=n&&dee==0;i++){
			int num=a[i].length();
			for(int j=leg;j<num-1&&dee==0;j++){
				temp=a[i].substr(0,j);
				if(cha(temp)!=-1){
					cout<<"NO"<<endl;
					dee=1;
				}
			}
		}
		if(dee==0)cout<<"YES"<<endl;
	}
}