HDU-1251-统计难题

统计难题

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 54701    Accepted Submission(s): 19121

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

 

法一：字典树（占空间）

法二：map

#include <iostream>
#include <cstdio>
using namespace std;
const int kind = 26;
struct Treenode
{
    size_t cnt;
    Treenode* next[kind];
    Treenode()
    {
        cnt = 1;
        for(int i=0;i<kind;i++)
            next[i] = NULL;
    }
};
void insert(Treenode *&root,char* word)
{
    Treenode *location = root;
    int i=0,branch=0;
    if(location==NULL)
    {
        location = new Treenode();
        root = location;
    }
    while(word[i])
    {
        branch = word[i] - 'a';
        if(location->next[branch])
            location->next[branch]->cnt++;
        else
        {
            location->next[branch] = new Treenode();
        }
        i++;
        location = location->next[branch];
    }
}
int search(Treenode *&root,char* word)
{
    Treenode* location = root;
    int i=0,branch = 0,ans=0;
    while(word[i])
    {
        branch = word[i]-'a';
        if(!location->next[branch])
            return 0;
        else
        {
            i++;
            location = location->next[branch];
            ans = location->cnt;
        }
    }
    return ans;
}
int main()
{
    char word[11];
    char ask[11];
    Treenode *root = NULL;
    while(gets(word))
    {
        if(word[0]=='')
            break;
        insert(root,word);
    }
    while(gets(ask))
        cout<<search(root,ask)<<endl;
    return 0;
}

#include <iostream>
#include <map>
#include <cstdio>
#include <string.h>
using namespace std;
int main()
{
    char a[11],b[11];
    map<string,int> cnt;

    while(gets(a))
    {
        if(a[0]=='')
            break;
        int len = strlen(a);
        for(int i=0;i<len;i++)
        {
            int j;
            for(j =0;j<=i;j++)
            {
                b[j]=a[j];
            }
            b[j] = '';
            string str(b);
            cnt[str]++;
        }
    }
    while(gets(b))
    {
        cout<<cnt[b]<<endl;
    }

    return 0;
}