UVALive - 3942 Remember the Word(trie + dp)

題目鏈接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=22109；

思路：dp[i]表示原串從i開始的後段子串的方案數，從後往前dp，然後dp[i] = sum(dp[i + len(x)]，這裏的x是給出單詞並且是這後段的前綴串。如果直接枚舉的話會超時，先把單詞全部加到字典樹上面然後在字典樹上面查找就可以了。我是用指針的寫法，記住每組數據輸出後一定要初始化字典樹！

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int inf = 0x7fffffff, N = 3e5 + 10, mo = 20071027;
typedef long long ll;
using namespace std;
struct trie
{
    int data;
    trie *Next[26];
    trie()
    {
        data = 0;
        memset(Next, 0, sizeof(Next));
    }
};
char s[N], x[110];
int dp[N], len;
trie *root;
void trie_insert(char *str)
{
    trie *p = root;
    for(int i = 0; str[i]; i++)
    {
        int ch = str[i] - 'a';
        if(p->Next[ch] == NULL) p->Next[ch] = new trie;
        p = p->Next[ch];
    }
    p->data++;
}
int trie_find(char *str, int sta)
{
    int cnt = 0;
    trie *p = root;
    for(int i = 0; str[i]; i++)
    {
        int ch = str[i] - 'a';
        p = p->Next[ch];
        if(p == NULL) return cnt;
        if(p->data) cnt = (cnt + dp[i + sta + 1])%mo;
    }
    return cnt;
}
void del(trie * p)
{
    if(p == NULL)
        return ;
    else
    {
        for(int i = 0; i<26; i++)
            del(p->Next[i]);
    }
    delete p;
}
int main()
{
    int n, cas = 1;

    while(~scanf("%s", s))
    {
        root = new trie;
        scanf("%d", &n);
        memset(dp, 0, sizeof(dp));
        len = strlen(s);
        dp[len] = 1;
        for(int i = 1; i<=n; i++)
        {
            scanf("%s",x);
            trie_insert(x);
        }
        for(int i = len - 1; i>=0; i--)
            dp[i] = trie_find(s+i, i);
        printf("Case %d: %d\n", cas++, dp[0]);
        del(root);
    }
    return 0;
}