20140723 【字典树 - 前缀树，指针版】 HDOJ 1251 统计难题

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【BestCoder Round #2 来了！】7月27号19:00~21:00（赛前30分钟停止注册比赛）

统计难题 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others) Total Submission(s): 17510 Accepted Submission(s): 7570 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 banana band bee absolute acm ba b band abc Sample Output 2 3 1 0 Author Ignatius.L Recommend Ignatius.L | We have carefully selected several similar problems for you: 1075 1247 1671 1298 1800 Statistic | Submit | Discuss | Note

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#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
#define MAX_son 26

struct Node{
int en;
int num;
Node *next[MAX_son];
Node(){
memset(next, 0, sizeof(next));
en = 0;
num = 0;
}
};

Node *root = new Node;

int idx(char *s){
return *s - 'a';
}

void Insert(char *s){
Node *p = root, *np;
while( *s ){
int c = idx(s);
if( NULL == p->next[c] ){
np = new Node;
p->next[c] = np;
}
p->next[c]->num++;
p = p->next[c];
s++;
}
p->en = 1;
}

int Find(char *s){
Node *p=root, *np;
while( *s ){
int c = idx(s);
if( NULL == p->next[c] )    return 0;
p = p->next[c];
s++;
}
return p->num;
}

void del(Node *p){
for(int i=0; i<MAX_son; i++)
if( NULL != p->next[i] )
del( p->next[i] );
delete p;
}

int main(){    // freopen("1251.txt", "r", stdin);

char s[20];
while( cin.getline(s, 18) ){   // 读空行真是一个悲伤的故事
if( s[0]=='\0' )    break;
Insert(s);
}
while( EOF != scanf("%s", s) ){
printf("%d\n", Find(s) );
}
del(root);
return 0;
}