POJ - 3630 Phone List (字典树)

POJ - 3630 Phone List (字典树)

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

题目大意：

给你一些电话号码。
问是否有一个电话号码是另一个的前缀。

解题思路：

设一个sum数组，sum[i] 表示i这个节点被经过了几次。
flag[i]表示标号为i的是否为一个字符串的最后一个
然后插入每个单词后
遍历所有的节点 如果flag[i]==1并且sum[i]>=2 就说明 这个点是为结尾的字母，并且这个点被经过了两次，说明以这个节点结尾的电话号码一定是另一个的前缀。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
const int maxn = 2e5+10;
int tree[maxn][11];//30取决于字母的个数,第i个节点的第j个孩子
int tot = 0;
int sum[maxn];
bool flag[maxn];//标号为i的是否为一个字符串的最后一个
void add(char *s){
	int root = 0;
	int id;
	int len = strlen(s);
	for(int i=0;i<len;i++){
		id = s[i]-'0';//这个字母是几
		if(!tree[root][id]) tree[root][id] = ++tot;
		root = tree[root][id];
		sum[root]++;
	}
	flag[root] = true;//这是个字符串
}
bool find(char *s){
	int root = 0;
	int id;
	ll res = 0;
	int len = strlen(s);
	for(int i=0;i<len;i++){
		id = s[i]-'a';
		root = tree[root][id];
		if(flag[root]){
			return false;
		}
	}
	return true;
}
char ss[maxn][20];
int n;
void init(){
	memset(tree,0,sizeof(tree));
	memset(flag,false,sizeof(flag));
	memset(sum,0,sizeof(sum));
	tot=0;
}
int main(){
	int t;
	cin>>t;
	while(t--){
		init();
		scanf("%d",&n);
		for(int i=0;i<n;i++){
			scanf("%s",ss[i]);
			add(ss[i]);
		
		}
		int tmp = 1;
		for(int i=1;i<=tot;i++){
			if(flag[i]&&sum[i]>=2){
				tmp = 0;
				break;
			}
		}
		if(tmp) printf("YES\n");
		else printf("NO\n");
	

	}
	return 0;
}