UVA3942 Remember the Word

题意：输入一个字符串(l<300000)，一个n(n<4000)，接下来n个字符串(l<100)，问能分解成几种字符串

题解：直接计算复杂度高，考虑DP，dp[i]代表以第i个字符开头的字符串分解的方案数，dp[i] = sum(dp[j+len])，len代表可以找到的字符串长度，多模式的字符串用到了char树，放在char树上跑一遍，找到以i+1开头的字符串复杂度,一开始写搜索。。后来发现复杂度过不去

#include <bits/stdc++.h>
#define ll long long
#define maxn 100*4000+10
#define siz 26
#define mod 20071027
using namespace std;
char s[300010], s1[110];
ll l, dp[300010];
struct Tire{
    ll ch[maxn][siz];
    ll val[maxn], sz;//节点总数
    Tire(){sz = 1;memset(ch, 0, sizeof(ch));}
    ll idx(char c){return c-'a';}
    void insert(char *s, ll v){
        ll u = 0,n = strlen(s);
        for(ll i=0;i<n;i++){
            ll 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;
    }
}Tir;
int main(){
    ll n, t = 0;
    while(~scanf("%s", s)){
        l = strlen(s);
        memset(Tir.ch, 0, sizeof(Tir.ch));
        Tir.sz = 1;
        memset(dp, 0, sizeof(dp));
        dp[l] = 1;
        scanf("%lld", &n);
        for(ll i=1;i<=n;i++){
            scanf("%s", s1);
            Tir.insert(s1, i);
        }
        for(int i=l-1;i>=0;i--){
            int u = 0;
            for(int j=i;j<l;j++){
                if(Tir.ch[u][s[j]-'a']==0) break;
                u = Tir.ch[u][s[j]-'a'];
                if(Tir.val[u]) dp[i] = (dp[i]+dp[j+1])%mod;
            }
        }
        printf("Case %lld: %lld
", ++t, dp[0]);
    }
    return 0;
}