【字典树】POJ_3630_Phone List

Phone List
Time Limit: 1000MS
Memory Limit: 65536K

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

题意：
给n个电话簿，每个电话簿中有m个电话号码，如果电话簿中有某一个号码是无效的（即另一个号码为其前缀），则输出NO，如果没有则输出YES。

思路：
字典树。

#include<cstdio>
#include<cstring>
using namespace std;

char phone[10 + 5];
bool yes;

int cnt;
struct Node {
    int next[10];
    int has;
}node[1000000 + 10];

void build(int root, char* str, bool hasNew) {
    int th = str[0] - '0';
    if (node[root].next[th] == 0) {
        node[root].next[th] = cnt++;
        hasNew = true;
    }
    if (node[node[root].next[th]].has == 1)
        yes = false;
    if (str[1] == '\0') {
        node[node[root].next[th]].has = 1;
        if (!hasNew)
            yes = false;
        return;
    }
    build(node[root].next[th], str + 1, hasNew);
    return;
}

int main() {
    int T, n;
    scanf("%d", &T);
    while (T--) {
        memset(node, 0, sizeof(node));
        cnt = 2;
        yes = true;
        scanf("%d", &n);
        for (int i = 0; i < n; ++i) {
            scanf("%s", phone);
            build(1, phone, false);
        }
        if (yes)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}