zoukankan      html  css  js  c++  java
  • LA 3942 Remember the Word 字典树 dp

    Neal is very curious about combinatorial problems, and now here comes a problem about words. Knowing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie.

    Since Jiejie can't remember numbers clearly, he just uses sticks to help himself. Allowing for Jiejie's only 20071027 sticks, he can only record the remainders of the numbers divided by total amount of sticks.

    The problem is as follows: a word needs to be divided into small pieces in such a way that each piece is from some given set of words. Given a word and the set of words, Jiejie should calculate the number of ways the given word can be divided, using the words in the set.

    Input 

    The input file contains multiple test cases. For each test case: the first line contains the given word whose length is no more than 300 000.

    The second line contains an integer S , 1$ \le$S$ \le$4000 .

    Each of the following S lines contains one word from the set. Each word will be at most 100 characters long. There will be no two identical words and all letters in the words will be lowercase.

    There is a blank line between consecutive test cases.

    You should proceed to the end of file.

    Output 

    For each test case, output the number, as described above, from the task description modulo 20071027.

    Sample Input 

    abcd 
    4 
    a 
    b 
    cd 
    ab
    

    Sample Output 

    Case 1: 2

    ------------------------------------------------------------
    给出一个由S个不同单词组成的字典和一个长字符串,把这个字符串分解成若干个单词的链接,单词可以重复使用,有多少种方法?
    f[i]表示以字符i为结尾的单词分解方法,若一个长度为x的单词与以i+1为开头的部分相匹配,则f[i+x]=f[i+x]+f[i];
    将单词建立一个字典树,用字典树快速求出匹配的单词。


    ------------------------------------------------------------
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    
    using namespace std;
    
    char str[333333];
    char words[4444][111];
    int f[333333];
    
    //用类,或者结构体定义都行
    struct trie
    {
        trie* next[26];
        int num;
        int value;
        trie()
        {
           for(int i=0;i<26;i++) next[i]=NULL;
           value=0;//记录是不是一个单词
           num=0;//记录单词出现的次数
        }
        void clear()
        {
           for(int i=0;i<26;i++) next[i]=NULL;
           value=0;//记录是不是一个单词
           num=0;//记录单词出现的次数
        }
    }root;
    
    //插入:
    void insert(char* s)
    {
        trie* p=&root;
        int k=0;
        while(s[k]!='\0')
        {
            if(!p->next[s[k]-'a']) p->next[s[k]-'a']=new trie;
            p=p->next[s[k]-'a'];
            p->num++;
            k++;
        }
        p->value=1;
    }
    
    //查找
    void find(char* s,int pos)
    {
        trie* p=&root;
        int k=0;
        while(s[k]!='\0'&&p->next[s[k]-'a'])
        {
            p=p->next[s[k]-'a'];
            if (p->value==1)
            {
                f[pos+k+1]=(f[pos+k+1]+f[pos])%20071027;
            }
            k++;
        }
    }
    
    int main()
    {
        int l,s;
        int cnt=0;
        while (~scanf("%s",str+1))
        {
            cnt++;
            memset(f,0,sizeof(f));
            root.clear();
            scanf("%d",&s);
            l=strlen(str+1);
            for (int i=0;i<s;i++)
            {
                scanf("%s",words[i]);
                insert(words[i]);
            }
            f[0]=1;
            for (int i=1;i<=l;i++)
            {
                if (f[i-1])
                {
                    find(str+i,i-1);
                }
            }
            printf("Case %d: %d\n",cnt,f[l]);
        }
        return 0;
    }
    








  • 相关阅读:
    mysql 插入字段 字符串
    python---统计列表中数字出现的次数
    pycharm激活
    Jmeter遇到的坑
    通用接口测试用例设计【转】
    selenium系列------元素定位套路
    jmeter系列-------注意事项
    jmeter系列-------脚本编写格式
    jmeter系列------变量的提取
    jmeter系列------参数关联
  • 原文地址:https://www.cnblogs.com/cyendra/p/3038415.html
Copyright © 2011-2022 走看看