UVALive 3942 - Remember the Word 字典树+dp

状态转移方程很容易想出,和背包思想有点像,难点在如何如何用字典树来搞dp,ps：原来在建stuct时初始化时可以节省不少时间的！！

#include<cstdio>
#include<cstring>
using namespace std;
const int max_node=400100;
const int modn=20071027;
const int son_num=26;
int ch[max_node][son_num];
int val[max_node];
int sz;
int dp[300100];
struct trie
{
    trie()
    {
        sz=1;
        memset(ch,0,sizeof(ch));
        memset(val,0,sizeof(val));
    }
    int idx(char c)
    {
        return c-'a';
    }
    void insert(char *s)
    {
        int u,c,n,i,j;
        u=0;
        n=strlen(s);
        for(i=0;i<n;i++)
        {
            c=idx(s[i]);
            if(!ch[u][c])ch[u][c]=sz++;
            u=ch[u][c];
        }
        val[u]=1;
    }
    int query(char *T,int pos)
    {
        int u,c,n,i,j;
        int res;
        u=0;
        n=strlen(T);
        res=0;
        for(i=pos;i<n&&i<=pos+100;i++)
        {
            c=idx(T[i]);
            if(!ch[u][c])return res;
            u=ch[u][c];
            if(val[u])res=(res+dp[i+1])%modn;
        }
        return res;
    }
};
char T[300100],s[110];
int main()
{
    int cas,n,i,j,len;
    cas=1;
    while(scanf("%s",T)!=EOF)
    {
        scanf("%d",&n);
        trie op;
        while(n--)
        {
            scanf("%s",s);
            op.insert(s);
        }
        len=strlen(T);
        dp[len]=1;
        for(i=len-1;i>=0;i--)
        dp[i]=op.query(T,i);
        printf("Case %d: %d\n",cas++,dp[0]);
    }
}