HDU 1671 Phone List

Problem 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:
1. Emergency 911
2. Alice 97 625 999
3. 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

大意是给出g组每组n个电话号码，判断该组号码是否合法（如91125426和911这两个号码不合法，拨出911会直接接通第二个号码，第一个号码打不过去）

由于从头开始找子串，用 黑科技 字典树。

每输入一个号码，逐位创建子节点，号码结束后设置布尔值标记。

用结构体建树：isend判断是否为单词结尾，next存储指定儿子（0-9）的编号，若没有则记为0。

一个比较暴力的算法：遍历一遍字典树，如果有一个节点是单词的结尾，但next不为空（有儿子），则输出NO。

#include <bits/stdc++.h>
using namespace std;
struct Node{
    bool isend;
    int next[10];
};
Node trieTree[100010];

int size=1;
void Insert(char c[]){
    int curr=1;
    int len=strlen(c);
    for(int i=0; i<len; i++){
        int num = c[i]-'0';
        if(!trieTree[curr].next[num]) trieTree[curr].next[num] = ++size;
        curr = trieTree[curr].next[num];
    }
    trieTree[curr].isend = true;
    return;
}
void init(){
    size=1;
    memset(trieTree, 0, sizeof(trieTree));
}
bool isempty(int a[]){
    for(int i=0; i<10; i++){
        if(a[i]) return false;
    }
    return true;
}

int main(){
    int g;
    cin >> g;
    while(g--){
        init();
        int n;
        cin >> n;
        while(n--){
            char str[20];
            cin >> str;
            Insert(str);
        }
        bool flag = true;
        for(int i=1; i<=size; i++){
            if(trieTree[i].isend && !isempty(trieTree[i].next)) flag = false;
        }
        if(flag) cout << "YES\n";
        else cout << "NO\n";
    }
    return 0;
}

稍微巧妙一点的算法是，在输入过程直接判断，若有节点已被标记为isend（该号码包含了其他号码）或整个输入过程没有创建新的节点（有其他号码包含该号码）则输出NO。

先记一下，回头放思路改良后的代码。