Phone List POJ - 3630 (字典树或string)

Phone List

POJ - 3630

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

方法1：

先从小打到排序再利用string中的substr进行比较

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
int main(){
    int t,n;
    string str[10010];
    string temp;
    int len;
    bool flag;
    scanf("%d",&t);
    while(t--){
        int i;
        scanf("%d",&n);
        flag = 1;
        for(i = 0; i < n; i++){
            cin >> str[i];
        }
        sort(str,str+n);
        for(i = 0; i < n-1; i++){
            len = str[i].length();
            temp = str[i+1].substr(0,len);
            if(temp == str[i]){
                flag = 0;
                break;
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

方法二

字典树静态建树，动态建树一直超时

code：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
struct Tree{
    int time;
    struct Tree* Next[10];
}T[100005];

int num = 0;

Tree* TreeNode(){
    Tree* p = &T[num++];
    p->time = 0;
    int i;
    for(i = 0; i < 10; i++)
        p->Next[i] = NULL;
    return p;
}

void Insert(Tree* &root,char number[]){
    int i = 0;
    if(root == NULL)
        root = TreeNode();
    Tree* p = root;
    while(number[i]){
        if(p->Next[number[i]-'0'] == NULL){
            p->Next[number[i]-'0'] = TreeNode();
        }
        p = p->Next[number[i]-'0'];
        p->time++;
        i++;
    }
}
bool judge(Tree* root,char number[]){
    int i = 0;
    if(root == NULL)
        return true;
    Tree* p = root;
    while(number[i]){
        p = p->Next[number[i]-'0'];
        if(p->time == 1)
            return false;
        i++;
    }
    return true;
}
int main(){
    bool flag = true;
    int i,j,n,t;
    Tree* root = NULL;
    char number[100005][11];
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        root = NULL;
        for(i = 0; i < n; i++){
            scanf("%s",number[i]);
            Insert(root,number[i]);
        }
        for(i = 0; i < n; i++){
            if(judge(root,number[i])){
                flag = false;
                break;
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
        flag = true;
        num = 0;
    }
    return 0;
}