【字典树】杭电1671 : Phone List

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5747    Accepted Submission(s): 1973

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

#include <stdio.h>

typedef struct NODE
{
    int flag; /* 标记是否是最后一个字母 */
    struct NODE *pNext[10];
}NODE,*PNODE;

NODE node[100100];
int ix_node = 0;

PNODE CreateNode()
{
    PNODE p = &node[ix_node++];
    int i;
    p->flag = 0; /* 不是最后一个字母 */

    for(i = 0 ; i < 10 ; ++i)
        p->pNext[i] = NULL;

    return p;
}

int Insert(PNODE *pRoot,char *s)
{
    PNODE p = *pRoot;
    int i,k;

    if(!p)
        p = *pRoot = CreateNode();

    i = 0;
    while(s[i])
    {
        k = s[i] - '0';

        if(s[i + 1]) /* 如果不是最后一个数字 */
        {
            if(!p->pNext[k])
            {
                p->pNext[k] = CreateNode();
            }
            else
            {
                if(1 == p->pNext[k]->flag) /* 有前缀了 */
                    return 0;
            }
        }
        else /* 当前数字已经是最后一个数了 */
        {
            if(p->pNext[k] != NULL) /* 当前串是前缀 */
                return 0;

            if(!p->pNext[k])
            {
                p->pNext[k] = CreateNode();
                p->pNext[k]->flag = 1; /* 设标记为1 */
            }
            else
            {
                if(1 == p->pNext[k]->flag) /* 有前缀了 */
                    return 0;
            }
        }

        p = p->pNext[k];
        ++i;
    }

    return 1;
}

void Delete(PNODE *pRoot)
{
    int i,j;
    for(i = 0 ; i < ix_node ; ++i)
    {
        for(j = 0 ; j < 10 ; ++j)
            node[i].pNext[j] = NULL;
    }

    ix_node = 0;
    *pRoot = NULL;
}

int main()
{
    int N,t;
    int IsOk = 1;
    char str[15];
    scanf("%d",&N);

    PNODE pRoot = NULL;

    while(N--)
    {
        scanf("%d",&t);
        while(t--)
        {
            scanf("%s",str);
            if(!Insert(&pRoot,str))
                IsOk = 0;
        }
        if(IsOk)
            printf("YES\n");
        else
            printf("NO\n");

        IsOk = 1;
        Delete(&pRoot);

    }

    return 0;
}