POJ3630 HDU1671 ZOJ2876 UVA11362 Phone List【字典樹】

Phone List

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 33225	Accepted: 9609

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

Nordic 2007

問題連結：POJ3630 HDU1671 ZOJ2876 UVA11362 Phone List

問題描述：

　　輸入若幹數字組成的字元串集合，判斷這些字元串是否一緻，若其中某個字元串是為另外一個字元串的字首則不是一緻的。

問題分析：

  該問題用建構字典樹的方法來解決。構造一個字典樹進行判斷即可。

程式說明：（略）

參考連結：（略）

題記：（略）

AC的C++語言程式如下：

/* POJ3630 HDU1671 ZOJ2876 UVA11362 Phone List */

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

const int N = 10000;
const int LEN = 10;
const int SIZE = 10;
const char SCHAR = '0';

struct Trie {
    int acnt;   // access count
    int ccnt;   // child number
    int childs[SIZE];
    void init()
    {
        acnt = 1;
        ccnt = 0;
        memset(childs, 0, sizeof(childs));
    }
} trie[N * LEN];
int ncnt;   // Trie Node count
char s[LEN + 1]; // Input

bool insert(char s[])
{
    int p = 0;
    for(int i = 0; s[i]; i++) {
        int k = s[i] - SCHAR;
        int child = trie[p].childs[k];
        if(child) {
            trie[child].acnt++;
            p = child;
        } else {
            if(i > 0 && trie[p].acnt > 1 && trie[p].ccnt == 0)
                return false;
            trie[++ncnt].init();
            trie[p].childs[k] = ncnt;
            trie[p].ccnt++;
            p = ncnt;
        }
    }
    return trie[p].ccnt == 0;
}

int query(char s[])
{
    int p = 0;
    for (int i = 0; s[i]; i++) {
        int k = s[i] - SCHAR;
        int child = trie[p].childs[k];
        if (child == 0)
            return 0;
        else
            p = child;
    }
    return trie[p].ccnt;
}

int main()
{
    int t, n;
    scanf("%d", &t);
    while(t--) {
        bool flag = true;
        ncnt = 0;
        trie[ncnt].init();

        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%s", s);
            if(flag)
                if(!insert(s))
                    flag = false;
        }

        printf("%s\n", flag ? "YES" : "NO");
    }

    return 0;
}