HDU 1671 （字典樹統計是否有首碼）

標籤：字典樹

題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1671

Problem DescriptionGiven 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.
 
InputThe 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. 
OutputFor each test case, output “YES” if the list is consistent, or “NO” otherwise. 
Sample Input

2391197625999911254265113123401234401234598346

 
Sample Output

NOYES

 
Source2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

題意：

統計是否有相同的首碼！

代碼如下：

#include <cstdio>#include <cstring>#include <malloc.h>#include <iostream>using namespace std;#define MAXN 10typedef struct Trie{    int v;//根據需要變化    Trie *next[MAXN];    //next是表示每層有多少種類的數，如果只是小寫字母，則26即可，    //若改為大小寫字母，則是52，若再加上數字，則是62了}Trie;Trie* root;void createTrie(char *str){    int len = strlen(str);    Trie *p = root, *q;    for(int i = 0; i < len; i++)    {        int id = str[i]-'0';        if(p->next[id] == NULL)        {            q = (Trie *)malloc(sizeof(Trie));            q->v = 1;//初始v==1            for(int j = 0; j < MAXN; j++)                q->next[j] = NULL;            p->next[id] = q;            p = p->next[id];        }        else        {            p->next[id]->v++;            p = p->next[id];        }    }     p->v = -1;//若為結尾，則將v改成-1表示}int findTrie(char *str){    int len = strlen(str);    Trie *p = root;    for(int i = 0; i < len; i++)    {        int id = str[i]-'0';        p = p->next[id];        if(p == NULL) //若為空白集，表示不存以此為首碼的串            return 0;        if(p->v == -1)   //字元集中已有串是此串的首碼            return -1;    }    //return p->v;    return -1;   //此串是字元集中某串的首碼}int dealTrie(Trie* T){    //動態字典樹，有時會超記憶體,這是就要記得釋放空間了    if(T==NULL)        return 0;    for(int i = 0; i < MAXN; i++)    {        if(T->next[i]!=NULL)            dealTrie(T->next[i]);    }    free(T);    return 0;}int main(){    int t, n;    char str[15];    scanf("%d",&t);    while(t--)    {        int flag = 0;        root = (Trie *)malloc(sizeof(Trie));        for(int i = 0; i < MAXN; i++)            root->next[i] = NULL;        scanf("%d",&n);        for(int i = 0; i < n; i++)        {            scanf("%s",str);            if(findTrie(str) == -1)            {                flag = 1;                continue;            }            createTrie(str);        }        if(flag)            printf("NO\n");        else                printf("YES\n");        dealTrie(root);    }    return 0;}

HDU 1671 （字典樹統計是否有首碼）