HDU 1671 && POJ 3630 Phone List (Trie树 好题)

Phone List

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

Total Submission(s): 11721    Accepted Submission(s): 3982

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）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1671

题目大意：有n个号码，如果某个是另一个的前缀就是NO，否则是YES

题目分析：字典树的好题，判断时候有两点：
1.如果一个单词插入字典树时没有p -> next[idx] == NULL的情况，就是说其是某个已经存在的单词的前缀，则flag标记
2.如果查找时，还没找到头，就有p -> end == true，则说明存在其前缀，flag标记
还有一点要注意的是这题用指针做得delete否则mle

#include 
#include 
char s[15];
bool flag;

struct node
{
    node *next[10];
    bool end;
    node()
    {
        memset(next, false, sizeof(next));
        end = false;
    }
};

void Insert(node *p, char *s)
{
    bool f = true;
    for(int i = 0; s[i] != '\0'; i++)
    {
        int idx = s[i] - '0';
        if(p -> next[idx] == NULL)
        {
            p -> next[idx] = new node();
            f = false;
        }
        p = p -> next[idx];
    }
    p -> end = true;
    if(f)
        flag = true;
}

void Search(node *p, char *s)
{   
    for(int i = 0; s[i] != '\0'; i++)
    {
        int idx = s[i] - '0';
        if(p -> end)
        {
            flag = true;
            return;
        }
        if(p -> next[idx] != NULL)
            p = p -> next[idx];
    }
}

void Delete(node *p)
{
    for(int i = 0; i < 10; i++)
        if(p -> next[i]) 
            Delete(p -> next[i]);
    delete p;
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        flag = false;
        node *root = new node();
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", s);
            Insert(root, s);
            if(!flag)
                Search(root, s);
        }
        printf("%s\n", flag ? "NO" : "YES");
        Delete(root);
    }
}

该题在poj上动态分配会T，要写静态的

#include 
#include 
char s[15];
bool flag;
int cnt = 0;

struct node
{
    node *next[10];
    bool end;
}tree[100005];

inline void Init(node *p)
{
    memset(p -> next, NULL, sizeof(p -> next));
    p -> end = false;
}
void Insert(node *p, char *s)
{
    bool tmp = true;
    for(int i = 0; s[i]; i++)
    {
        int idx = s[i] - '0';
        if(p -> next[idx] == NULL)
        {
            tmp = false;
            cnt++;
            p -> next[idx] = tree + cnt;
            Init(p -> next[idx]);
        }
        p = p -> next[idx];
    }
    p -> end = true;
    if(tmp)
        flag = true;
}

void Query(node *p, char *s)
{
    for(int i = 0; s[i]; i++)
    {
        int idx = s[i] - '0';
        if(p -> end)
        {
            flag = true;
            return;
        }
        if(p -> next[idx] != NULL)
            p = p -> next[idx];
    }
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        cnt = 0;
        flag = false;
        scanf("%d", &n);
        node *root = tree;
        Init(root);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", s);
            Insert(root, s);
            if(!flag)
                Query(root, s);
        }
        printf("%s\n", flag ? "NO" : "YES");
    }
}