POJ 3630 Phone List（字符串前缀重复）题解

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

思路：

树状数组超时。这里给每一串先排序，然后比较相邻的两个有没有前缀重复就可以了，有点厉害（所以为什么他会出现在树状数组专题呢）。

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<cmath>
#include<string>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<iostream>
#include<algorithm>
#include<sstream>
#define ll long long
const int N=10005;
const int INF=1e9;
using namespace std;
char s[N][15];
int cmp(const void *a,const void *b){
	return strcmp((char *)a,(char *)b);
}
int main(){
	int flag,t,n;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i=0;i<n;i++){
			scanf("%s",s[i]);
		}
		qsort(s,n,sizeof(s[0]),cmp);
		flag=0;
		for(int i=0;i<n-1;i++){
			if(strncmp(s[i],s[i+1],strlen(s[i]))==0){
				flag=1;
				break;
			}
		}
		if(flag) printf("NO\n");
		else printf("YES\n");
	}
    return 0;
}