HDU 1671 Phone List （Trie）

pid=1671">Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 12879    Accepted Submission(s): 4391

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

 

Source

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

解析：求是否为非前缀串（即随意串的前缀都不是一个已有的串）。在建Trie树时，仅仅标记串的最后一个字符，然后查询时，看每一个串的中间是否存在被标记的点就可以。

AC代码：

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

const int maxnode = 100000 * 10 + 5;
const int sigma_size = 10;

struct Trie{
    int ch[maxnode][sigma_size];
    int val[maxnode];
    int sz;

    void clear(){ sz = 1; memset(ch[0], 0, sizeof(ch[0])); }  //初始仅仅有一个根节点
    int idx(char c){ return c - '0'; }

    void insert(string s){
        int u = 0, n = s.size();
        for(int i=0; i<n; i++){
            int c = idx(s[i]);
            if(!ch[u][c]){
                memset(ch[sz], 0, sizeof(ch[sz]));
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        val[u] ++;    //仅仅标记串尾的字符
    }

    bool find(string s){
        int u = 0, n = s.size();
        for(int i=0; i<n; i++){
            int c = idx(s[i]);
            if(val[u]) return false;    //在一个串的中间出现了已经标记的点
            u = ch[u][c];
        }
        return true;
    }

};

Trie T;
vector<string> S;

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    int t, n;
    string s;
    scanf("%d", &t);
    while(t--){
        S.clear();
        T.clear();       //不要忘了初始化
        scanf("%d", &n);
        for(int i=0; i<n; i++){
            cin>>s;
            T.insert(s);
            S.push_back(s);
        }
        int cnt = S.size();
        int i;
        for(i=0; i<cnt; i++){
            if(!T.find(S[i])) break;
        }
        puts(i < cnt ?

"NO" : "YES");
    }
    return 0;
}