HDU - 1671 Phone List （字典树）

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本文介绍了一种算法，用于检查电话号码列表是否具有一致性，即任一号码都不应作为其他号码的前缀。通过构建特殊的树形数据结构来高效解决此问题。

Phone List

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21692 Accepted Submission(s): 7360

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

Sample Output

  
   NO
YES

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

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Statistic | Submit | Discuss | Note 题意：给出 n 个数字串，若它们互不为其它串的前缀输出 YES，否则输出 NO；

#include <bits/stdc++.h>
using namespace std;

int t, n;
char s[10001][41];
struct xx{
    int cnt;
    xx *nex[15];
    xx(){
        cnt = 0;
        memset(nex, NULL, sizeof nex);
    }
};

void Insert(xx *root, char a[]){
    int l = strlen(a);
    xx *p = root;
    for(int i = 0; i < l; i++){
        int id = a[i]-'0';
        if(p->nex[id] == NULL){
            p->nex[id] = new xx();
        }
        p = p->nex[id];
        (p->cnt)++;
    }
}

bool Query(xx *root, char a[]){
    int l = strlen(a);
    xx *p = root;
    for(int i = 0; i < l; i++){
        int id = a[i]-'0';
        p = p->nex[id];
    }
    for(int i = 0; i < 10; i++){
        if(p->nex[i] != NULL) return 0;
    }
    return 1;
}

void Delete(xx *root){
    xx *p = root;
    for(int i = 0; i < 10; i++){
        if(p->nex[i] != NULL){
            Delete(p->nex[i]);
        }
    }
    delete p;
}

int main(){
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        xx *root = new xx();
        for(int i = 0; i < n; i++){
            scanf("%s", s[i]);
            Insert(root, s[i]);
        }
        bool flag = 1;
        for(int i = 0; i < n; i++){
            flag = Query(root, s[i]);
            if(!flag) break;
        }
        printf(flag ? "YES\n" : "NO\n");
        Delete(root);
    }
}