山东第一届省赛1001 Phone Number（字典树）

Phone Number

 

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

We know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345, after pressing 123, we call A, and not able to call B.
Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.
 

输入

 The input consists of several test cases.
 The first line of input in each test case contains one integer N (0<N<1001), represent the number of phone numbers.
 The next line contains N integers, describing the phone numbers.
 The last case is followed by a line containing one zero.

输出

 For each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.

示例输入

2
012
012345
2
12
012345
0

示例输出

NO
YES

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

题意：给出几个字符串问是否会有前缀产生，直接字典树搞定

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int Max = 1000 + 10;
struct Node
{
    int value;
    int cnt;
    Node * Next[10];
};
Node * root;
char str[Max];
bool build(char * s)
{
    Node * p = root, *temp;
    int len = strlen(s);
    for (int i = 0; i < len; i++)
    {
        int id = str[i] - '0';
        if (p->Next[id] == NULL)
        {
            temp = new Node;
            temp->value = id;
            temp->cnt = 0;
            for (int i = 0; i < 10; i++)
                temp->Next[i] = NULL;
            p->Next[id] = temp;
        }
        p = p->Next[id];
        if (p->cnt)
            return false;
    }
    return true;
}
void destroy(Node * temp)
{
    if (temp == NULL)
        return;
    for (int i = 0; i < 10; i++)
        destroy(temp->Next[i]);
    delete temp;
}
int main()
{
    int n;
    while (scanf("%d", &n) != EOF && n)
    {
        root = new Node;
        for (int i = 0; i < 10; i++)
            root->Next[i] = NULL;
        bool flag = true;
        while (n--)
        {
            scanf("%s", str);
            if (!flag)
                continue;
            if (!build(str))
                flag = false;
        }
        if (flag)
            printf("YES\n");
        else
            printf("NO\n");
        destroy(root);
    }
    return 0;
}