hdu_1251 统计难题（字典树）

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统计难题

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 35641 Accepted Submission(s): 13324

Problem Description

Ignatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀).

Input

输入数据的第一部分是一张单词表,每行一个单词,单词的长度不超过10,它们代表的是老师交给Ignatius统计的单词,一个空行代表单词表的结束.第二部分是一连串的提问,每行一个提问,每个提问都是一个字符串.

注意:本题只有一组测试数据,处理到文件结束.

Output

对于每个提问,给出以该字符串为前缀的单词的数量.

Sample Input

bananabandbeeabsoluteacmbabbandabc

Sample Output

2310
字典树模板题
不过为了防止hdu爆栈，要加上一句#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 1000010#define mod 1000000007#define PI acos(-1.0)#define LL long long#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;struct Node{    Node *br[26];    int num;};Node *head;void initTrie(){    head = new Node;    for(int i = 0; i < 26; i++) head->br[i] = NULL;    head->num = 0;}void insertTrie(char *str){    Node *t, *s = head;    int len = strlen(str);    for(int i = 0; i < len; i++)    {        int id = str[i] - 'a';        if(s->br[id] == NULL)        {            t = new Node;            for(int j = 0; j < 26; j++)            {                t->br[j] = NULL;            }            t->num = 0;            s->br[id] = t;        }        s = s->br[id];        s->num++;    }}int findTrie(char *str){    Node *s = head;    int cot;    int len = strlen(str);    for(int i = 0; i < len; i++)    {        int id = str[i] - 'a';        if(s->br[id] == NULL)        {            return 0;        }        else        {            s = s->br[id];            cot = s->num;        }    }    return cot;}int main(){    initTrie();    char str[11];    while(gets(str) && str[0] != '\0') insertTrie(str);    while(gets(str)) printf("%d\n", findTrie(str));    return 0;}

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