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\n", ++t, dp[0]);}return 0;
}