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最新推荐文章于 2019-07-09 10:45:12 发布

原创最新推荐文章于 2019-07-09 10:45:12 发布 · 601 阅读

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#UVAlive

本文介绍了一个关于字符串组合的问题，即如何将一个给定的字符串通过一个预定义的单词集合进行分割，并计算出所有可能的分割方式的数量。文章提供了一段C++代码实现，该算法利用了前缀树（Trie）来高效地解决此问题。

Remember the Word

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

[Submit] [Go Back] [Status]

Description

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

http://vjudge.net/problem/viewProblem.action?id=22109

#include<cstdio>
#include<cstring>
using namespace std;
const int mode = 20071027;
const int maxn = 310000;
char str[maxn];
int dp[maxn];
char a[110];
struct node
{
    int next[26];
    int f;
} T[4000*100];
int tot;
void insert()
{
    int root=0;
    for(int i=0; a[i]; i++)
    {
        if(T[root].next[ a[i]-'a']==-1)
        {
            T[root].next[ a[i]-'a' ] = tot++;
        }
        root=T[root].next[a[i]-'a' ];
    }
    T[root].f=1;
}
int main()
{
    int n;
    int ca=1;
    while(scanf("%s",str+1)!=EOF)
    {
        scanf("%d",&n);
        memset(T,-1,sizeof(T));
        tot=1;
        while(n--)
        {
            scanf("%s",a);
            insert();
        }
        memset(dp,0,sizeof(dp));
        int len=strlen(str+1);
        dp[0]=1;
        str[0]=1;
        for(int i=0; str[i]; i++)
        {
            int  root=0;
            for(int j=1; str[i+j]; j++)
            {
                root=T[root].next[ str[i+j]-'a' ];
                if(root==-1 ) break;
                else
                {
                    if(T[root].f==1)
                    {
                        dp[i+j]+=dp[i];
                        while(dp[i+j]>mode)dp[i+j]-=mode;
                    }
                }

            }
             for(int i=1;str[i];i++)
            printf("%d ",dp[i]);
            puts("");
        }

        printf("Case %d: %d\n",ca++,dp[len]);
    }
    return 0;

}