LA 3942 Remember the Word

(Remember the Word ,LA 3942) 

题目来源：https://vjudge.net/problem/UVALive-3942

题意：给定一个字符串S以及n个单词，字符用这n个单词进行拆分，输出拆分的方案数。

思路：dp+字典树

可以先将这n个单词存储于字典树中，并记dp[i]:字符i开始的字符串的分解方案数量(即后缀S[i...end]),则动态转移方程为dp[i]=sum{dp[i+len(x)] | 单词x是S[i...end]的前缀}.

找S[i...end]的前缀可以在字典树中查找。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<vector>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<cstdio>
using namespace std;
#define N_MAX 4000+5
#define L_MAX 4000*100+5
#define INF 0x3f3f3f3f
#define MOD 20071027
typedef long long ll;
int n;
int dp[L_MAX];
int len[N_MAX];
char S[L_MAX];
struct Trie {
    int ch[L_MAX][26];
    int val[L_MAX];
    int sz;
    Trie() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }
    void init() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }
    int idx(char c) { return c - 'a'; }
    void insert(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]) {
                memset(ch[sz], 0, sizeof(ch[sz]));
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        val[u] = v;
    }
    //找长度为len的字符串s的前缀
    void find(char *s, int len, vector<int>&ans) {
        int u = 0;
        for (int i = 0; i < len; i++) {
            if (s[i] == '')break;
            int c = idx(s[i]);
            if (!ch[u][c]) break;
            u = ch[u][c];
            if (val[u] != 0)ans.push_back(val[u]);
        }
    }
};

Trie trie;//不要放到main()下

int main() {
    int Case = 1;
    while (scanf("%s%d", S, &n) !=EOF) {
        memset(dp, 0, sizeof(dp));
        trie.init();
        for (int i = 1; i <= n; i++) {
            char x[100];
            scanf("%s", x);
            len[i] = strlen(x);
            trie.insert(x, i);//将单词存入字典树,单词编号为i
        }
        int L = strlen(S);
        dp[L] = 1;
        for (int i = L - 1; i >= 0; i--) {
            vector<int>ans;//存储单词的编号
            trie.find(S + i, L - i, ans);
            for (int j = 0; j < ans.size(); j++) {
                dp[i] = (dp[i] + dp[i + len[ans[j]]]) % MOD;
            }
        }
        printf("Case %d: %d
", Case++, dp[0]);
    }
    return 0;
}