uva 1401 - Remember the Word(字典树+dp)

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题目链接：uva 1401 - Remember the Word

题目大意：给出一个由S个不同单词组成的字典和一个长字符串。吧这个字符串分解成若干个单词的连接，有多少种方法。

解题思路：dp(i) = sum{ dp(i+len(x))}, 单词x为S[i...L]的前缀。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;const int MOD = 20071027;const int maxnode = 300000;const int sigma_size = 30;struct Trie {    int sz;    int val[maxnode];    int next[maxnode][sigma_size];    Trie () { clear(); }    int idx (char ch) {    return ch - 'a'; }    void clear();    void insert (char *s, int v);    bool query (char *s);}tree;int dp[maxnode];char word[maxn], str[maxnode];void init () {    int n;    tree.clear();    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%s", word);        tree.insert(word, 1);    }}void solve (int x) {    int u;    int& ans = dp[x];    ans = u = 0;    for (int i = x; str[i]; i++) {        int v = str[i] - 'a';        if (tree.next[u][v] == 0)            return;        u = tree.next[u][v];        if (tree.val[u])            ans = (ans + dp[i+1]) % MOD;    }}int main () {    int cas = 1;    while (scanf("%s", str) == 1) {        init();        int n = strlen(str);        dp[n] = 1;        for (int i = n-1; i >= 0; i--)            solve(i);        printf("Case %d: %d\n", cas++, dp[0]);    }    return 0;}void Trie::clear () {    sz = 1;    memset(next[0], 0, sizeof(next[0]));}void Trie::insert (char* s, int v) {    int u = 0, n = strlen(s);    for (int i = 0; i < n; i++) {        int c = idx(s[i]);        if (next[u][c] == 0) {            val[sz] = 0;            memset(next[sz], 0, sizeof(next[sz]));            next[u][c] = sz++;        }        u = next[u][c];    }    val[u] = v;}bool Trie::query (char* s) {    int u = 0, n = strlen(s);    for (int i = 0; i < n; i++) {        int c = idx(s[i]);        if (next[u][c] == 0)            return false;        u = next[u][c];    }    return val[u];}

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