UVALive - 3942 Remember the Word（字典树+dp）

Remember the Word

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit 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

S

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

Source

Regionals 2007 >> Asia
- Nanjing

分析：

大白书题目。字典树+dp，初学是很难的题目。好好理解！

ac代码：

#include <iostream>

#include<cstdio>

#include<cstring>

using namespace std;

#define MAXN 400000

const int mod=20071027;

struct node

{

int ch[27];

int v; // 0表示不是单词，其余数字表示为一个单词。

node()

{

memset(ch,0,sizeof(ch));

v=0;

}

}tree[MAXN];

int tot;

char op[300010];

char word[100];

int d[300010];

void insert(int v)

{

int len=strlen(word);

int root=0;

for(int i=0;i<len;i++)

{

if(tree[root].ch[word[i]-'a']==0) tree[root].ch[word[i]-'a']=tot++;

root=tree[root].ch[word[i]-'a'] ;

}

tree[root].v=v;

}

void cheak(int i,int len)

{

int ans=0;

int root=0;

for(int j=i;j<=len;j++)

{

if(tree[root].ch[op[j]-'a']==0) break;

root=tree[root].ch[op[j]-'a'];

if(tree[root].v) ans+=d[j+1]%mod;//dp思想

ans%=mod;

}

d[i]=ans;

}

int main()

{

int kase=1;

while(~scanf("%s",op))

{

memset(tree,0,sizeof(tree));

int len=strlen(op);

tot=1;

int s;

scanf("%d",&s);

for(int i=1;i<=s;i++)

{

scanf("%s",word);

insert(i);

}

memset(d,0,sizeof(d));

d[len]=1; // 初始化边界

for(int i=len-1;i>=0;i--)

{

cheak(i,len-1);

}

printf("Case %d: %d\n",kase++,d[0]%mod);

}

}

/**

abcd

4

a

b

cd

ab

abcd

0

**/