hdu 1251

统计拼图

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 16960    Accepted Submission(s): 7304

Problem Description

Ignatius近期遇到一个难题,老师交给他非常多单词(仅仅有小写字母组成,不会有反复的单词出现),如今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,每一个提问都是一个字符串.

注意:本题仅仅有一组測试数据,处理到文件结束.

 

Output

对于每一个提问,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

A字典树的第一道题

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
using namespace std;

struct edge{
    int u , v , letter , sum;
    edge(int a = 0 , int b = 0 , int c = 0 , int d = 0){
        u = a , v = b , letter = c , sum = d;
    }
};
vector<edge> e;
vector<int> head , next;

void add(int  u , int v , int letter , int sum){
    e.push_back(edge(u , v , letter , sum));
    next.push_back(head[u]);
    head[u] = next.size()-1;
    head.push_back(-1);
}

void addLibrary(string s){
    int u = 0 , index = 0;
    while(index < s.length()){
        int Next = head[u];
        while(Next != -1){
            if(e[Next].letter == s[index]-'a') break;
            Next = next[Next];
        }
        if(Next == -1) break;
        e[Next].sum++;
        index++;
        u = e[Next].v;
    }
    while(index < s.length()){
        add(u , head.size() , s[index]-'a' , 1);
        index++;
        u = head.size()-1;
    }
}

int query(string s){
    int u = 0 , index = 0 , Next;
    while(index < s.length()){
        Next = head[u];
        while(Next != -1){
            if(e[Next].letter == s[index]-'a') break;
            Next = next[Next];
        }
        if(Next == -1) return 0;
        index++;
        u = e[Next].v;
    }
    return e[Next].sum;
}

int main(){
    string dic;
    head.push_back(-1);
    while(getline(cin , dic)){
        if(dic.length() == 0) break;
        addLibrary(dic);
    }
    while(cin >> dic){
        printf("%d
" , query(dic));
    }
    return 0;
}