Ultraviolet A 1401 Remember the Word (Trie + DP), 1401 trie

Ultraviolet A 1401 Remember the Word (Trie + DP), 1401 trie

Question: n (n <= 4000) words and a string (len <= 300000) are given. How many methods are there to splice the string into several words?

Idea: Trie: create a word as Trie, and then run "dp" and "dp [I]" to indicate the situation where the I-th letter in the string starts, then, for each State, you only need to find the word starting with the corresponding letter on the Trie tree, and then dp [I] = sum {dp [I + len (x)]} after status transfer, x is one of n words and can be queried using Trie.

Code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 300005;const int MOD = 20071027;int ch[maxn][26];int val[maxn];int sz;int idx(char c){return c-'a';}void insert(char* s){    int u = 0;    int len = strlen(s);    for(int i = 0; i < len; i++){        int 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] = 1;}void init(){    sz = 1;    memset(ch[0],0,sizeof(ch[0]));    int m;    scanf("%d",&m);    char str[105];    while(m--){        scanf("%s",str);        insert(str);    }}int d[maxn];void find(int id, char* s){    d[id] = 0;    int u = 0;//    int len = strlen(s);    for(int i = id; s[i]; i++){        int c = idx(s[i]);        if(!ch[u][c]) return;        u = ch[u][c];        if(val[u]) d[id] = (d[id]+d[i+1])%MOD;    }}void solve(char* s){    int len = strlen(s);    d[len] = 1;    for(int i = len-1; i >= 0; i--){        find(i, s);    }    printf("%d\n",d[0]);}int main(){    int cas = 0;    char s[maxn];    while(scanf("%s",s) != EOF){        init();        printf("Case %d: ",++cas);        solve(s);    }    return 0;}

Note:

Do not use strlen () in the find function; otherwise, T_T is timed out .. Because the length of each call is obtained, and the time complexity is 0 (N * N );