LA 3942 ——Trie （前缀树）、DP

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 using namespace std;
 const int M = ;
 const int maxn =  + ;
 const int maxnode = +;
 const int _size = ;

 struct Trie {
     int ch[maxnode][_size];
     int val[maxnode];
     int sz;
     Trie() {
         sz = ;
         memset(ch[], , sizeof (ch[]));
     }
     int idx(int c) {return c - 'a';}

     void insert(char *s, int v) {   //插入字符串s和权值v
         int u = , n = strlen(s);
         for(int i = ; i < n; i++) {
             int c = idx(s[i]);
             if(!ch[u][c]) {
                 memset(ch[sz], , sizeof (ch[sz]));
                 val[sz] = ;
                 ch[u][c] = sz++;
             }
             u = ch[u][c];
         }
         val[u] = v;
     }
     void find_prefixes(const char *s, int len, vector<int>& ans) { //寻找字符串s的前缀
         int u = ;
         for(int i = ; i < len; i++) {
             if(s[i] == '\0') break;
             int c = idx(s[i]);
             if(!ch[u][c]) break;
             u = ch[u][c];
             if(val[u] != ) ans.push_back(val[u]);
         }
     }

     void clear() {
         sz = ;
         memset(val,  , sizeof val);
         memset(ch[], , sizeof (ch[]));
     }
 };

 Trie T;
 char S[maxn], s[];
 int d[maxn];

 int cal_d() {
     int L = strlen(S);
     d[L] = ;
     for(int i = L-; i >= ; i--){
         vector<int> p;
         T.find_prefixes(S+i, L-i, p);
         for(int j = ; j < p.size(); j++)  {
             d[i] = (d[i] + d[i + p[j]]) % M;
         }
     }
     return d[];
 }
 int main(int argc, const char * argv[]) {
     int tt = ;
     while(scanf("%s", S) == ){
         T.clear();
         memset(d, , sizeof d);
         int m;
         scanf("%d", &m);
         for(int i = ; i < m; i++) {
             scanf("%s", s);
             T.insert(s, strlen(s));
         }
         int ans = cal_d();
         printf("Case %d: %d\n", ++tt, ans);
     }
     return ;
 }

解题思路：

　　令d（i）表示从字符i开始的字符串(即S[i...L])的分解方案数。

　　则状态转移方程：

　　d(i) = sum{d(i+len(x)) | x为S[i..L]的前缀}