3942 - Remember the Word

Posted 2020-08-11 SJY

3942 - Remember the Word

思路：字典树+dp

dp[i]前i个字符，能由给的字串组成的方案数，那么dp[i] = sum(dp[i-k]);那么只要只要在字典树中查看是否有字串str[i-k+1,i]就行了；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<stdlib.h>
  4 #include<queue>
  5 #include<iostream>
  6 #include<string.h>
  7 #include<math.h>
  8 using namespace std;
  9 typedef long long LL;
 10 struct node
 11 {
 12     node *p[26];
 13     bool flag;
 14     node()
 15     {
 16         memset(p,0,sizeof(p));
 17         flag = false;
 18     }
 19 };
 20 char str[300005];
 21 char ans[300];
 22 char bns[300];
 23 node *head;
 24 void in(char *c,int l);
 25 void ask(char *c,int k,int l);
 26 LL dp[300005];
 27 const LL mod = 20071027;
 28 void fr(node *q);
 29 int main(void)
 30 {
 31     int i,j;
 32     int __ca = 0;
 33     while(scanf("%s",str)!=EOF)
 34     {
 35         int l = strlen(str);
 36         int n;
 37         scanf("%d",&n);
 38         head = new node();
 39         while(n--)
 40         {
 41             scanf("%s",ans);
 42             int k = strlen(ans);
 43             in(ans,k);
 44         }
 45         memset(dp,0,sizeof(dp));
 46         dp[0] = 1;
 47         for(i = 0; i < l; i++)
 48         {
 49             int x = max(0,i-100);
 50             int v = 0;
 51             for(j = i; j >= x; j--)
 52             {
 53                 bns[v++] = str[j];
 54             }
 55             bns[v] = ‘\0‘;
 56             ask(bns,i+1,v);
 57         }
 58         fr(head);
 59         printf("Case %d: ",++__ca);
 60         printf("%lld\n",dp[l]);
 61     }
 62     return 0;
 63 }
 64 void in(char *c,int l)
 65 {
 66     int i,j;
 67     node *root = head;
 68     for(i = l-1; i >=0; i--)
 69     {
 70         int x = c[i]-‘a‘;
 71         if(root->p[x]==NULL)
 72         {
 73             root->p[x] = new node();
 74         }
 75         root = root->p[x];
 76     }
 77     root->flag = true;
 78 }
 79 void ask(char *c,int k,int l)
 80 {
 81     int i,j;
 82     node *root = head;
 83     for(i = 0; i < l; i++)
 84     {
 85         int x = c[i]-‘a‘;
 86         if(root->p[x]==NULL)
 87         {
 88             return ;
 89         }
 90         else if(root->p[x]->flag)
 91         {
 92             dp[k] = dp[k] + dp[k-i-1];
 93             dp[k]%=mod;
 94         }
 95         root = root->p[x];
 96     }
 97 }
 98 void fr(node *q)
 99 {
100     int i,j;
101     for(i = 0; i < 26; i++)
102     {
103         if(q->p[i])
104         {
105             fr(q->p[i]);
106         }
107     }
108     free(q);
109 }