首页 > 代码库 > Uva1401(字典树)
Uva1401(字典树)
1401 - Remember the Word
Time limit: 3.000 seconds
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 <tex2html_verbatim_mark>, 1
S4000 <tex2html_verbatim_mark>.
Each of the following S <tex2html_verbatim_mark>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
下面是《算法竞赛入门经典--训练指南》代码仓库里的标程
// LA3942 Remember the Word// Rujia Liu#include<cstring>#include<vector>using namespace std;const int maxnode = 4000 * 100 + 10;const int sigma_size = 26;// 字母表为全体小写字母的Triestruct Trie { int ch[maxnode][sigma_size]; int val[maxnode]; int sz; // 结点总数 void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); } // 初始时只有一个根结点 int idx(char c) { return c - ‘a‘; } // 字符c的编号 // 插入字符串s，附加信息为v。注意v必须非0，因为0代表“本结点不是单词结点” void insert(const char *s, int v) { int u = 0, n = strlen(s); for(int i = 0; i < n; i++) { int c = idx(s[i]); if(!ch[u][c]) { // 结点不存在 memset(ch[sz], 0, sizeof(ch[sz])); val[sz] = 0; // 中间结点的附加信息为0 ch[u][c] = sz++; // 新建结点 } u = ch[u][c]; // 往下走 } val[u] = v; // 字符串的最后一个字符的附加信息为v } // 找字符串s的长度不超过len的前缀 void find_prefixes(const char *s, int len, vector<int>& ans) { int u = 0; for(int i = 0; i < len; i++) { if(s[i] == ‘\0‘) break; int c = idx(s[i]); if(!ch[u][c]) break; u = ch[u][c]; if(val[u] != 0) ans.push_back(val[u]); // 找到一个前缀 } }};#include<cstdio>const int maxl = 300000 + 10; // 文本串最大长度const int maxw = 4000 + 10; // 单词最大个数const int maxwl = 100 + 10; // 每个单词最大长度const int MOD = 20071027;int d[maxl], len[maxw], S;char text[maxl], word[maxwl];Trie trie;int main() { int kase = 1; while(scanf("%s%d", text, &S) == 2) { trie.clear(); for(int i = 1; i <= S; i++) { scanf("%s", word); len[i] = strlen(word); trie.insert(word, i); } memset(d, 0, sizeof(d)); int L = strlen(text); d[L] = 1; for(int i = L-1; i >= 0; i--) { vector<int> p; trie.find_prefixes(text+i, L-i, p); for(int j = 0; j < p.size(); j++) d[i] = (d[i] + d[i+len[p[j]]]) % MOD; } printf("Case %d: %d\n", kase++, d[0]); } return 0;}