Codeforces 271D - Good Substrings [字典树]

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D. Good Substrings

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You've got string s, consisting of small English letters. Some of the English letters are good, the rest are bad.

A substring s[l...r] (1 ≤ l ≤ r ≤ |s|) of string s  =  s₁s₂...s_|s| (where |s| is the length of string s) is string  s_ls_l + 1...s_r.

The substring s[l...r] is good, if among the letters  s_l, s_l + 1, ..., s_r there are at most k bad ones (look at the sample's explanation to understand it more clear).

Your task is to find the number of distinct good substrings of the given string s. Two substrings s[x...y] and s[p...q] are considered distinct if their content is different, i.e. s[x...y] ≠ s[p...q].

Input

The first line of the input is the non-empty string s, consisting of small English letters, the string's length is at most 1500 characters.

The second line of the input is the string of characters "0" and "1", the length is exactly 26 characters. If the i-th character of this string equals "1", then the i-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on.

The third line of the input consists a single integer k (0 ≤ k ≤ |s|) — the maximum acceptable number of bad characters in a good substring.

Output

Print a single integer — the number of distinct good substrings of string s.

Sample test(s)

Input

ababab
01000000000000000000000000
1

Output

5

Input

acbacbacaa
00000000000000000000000000
2

Output

8

Note

In the first example there are following good substrings: "a", "ab", "b", "ba", "bab".

In the second example there are following good substrings: "a", "aa", "ac", "b", "ba", "c", "ca", "cb".

 

9813017

2015-02-13 06:13:52

njczy2010

271D - Good Substrings

GNU C++

Accepted

248 ms

53000 KB

 

 

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<stack>
 11 #include<string>
 12 
 13 #define N 2250005
 14 #define M 1505
 15 #define mod 10000007
 16 //#define p 10000007
 17 #define mod2 1000000000
 18 #define ll long long
 19 #define LL long long
 20 #define eps 1e-6
 21 #define inf 100000000
 22 #define maxi(a,b) (a)>(b)? (a) : (b)
 23 #define mini(a,b) (a)<(b)? (a) : (b)
 24 
 25 using namespace std;
 26 
 27 int n;
 28 int T;
 29 int flag;
 30 int ccount;
 31 char ss[M];
 32 char f[30];
 33 
 34 typedef struct
 35 {
 36     int v;
 37     vector<int>nt;
 38 }PP;
 39 
 40 PP p[N];
 41 int le;
 42 
 43 int k;
 44 int tot;
 45 
 46 void insert(char s[])
 47 {
 48     int l=strlen(s);
 49     int i;
 50     int ff;
 51     int st=0;
 52     int y;
 53     vector<int>::iterator it;
 54     for(i=0;i<l;i++){
 55         ff=0;
 56         for(it=p[st].nt.begin();it!=p[st].nt.end();it++){
 57             y=*it;
 58             if(p[y].v==s[i]-'a'){
 59                 ff=1;
 60                 st=y;
 61                 break;
 62             }
 63         }
 64         if(ff==0){
 65             p[st].nt.push_back(tot);
 66             p[tot].v=s[i]-'a';
 67             p[tot].nt.clear();
 68             st=tot;
 69             tot++;
 70         }
 71     }
 72 }
 73 
 74 void ini()
 75 {
 76     ccount=0;
 77     int i;
 78     scanf("%s",f);
 79     scanf("%d",&k);
 80     flag=1;
 81     p[0].nt.clear();
 82     p[0].v=-1;
 83     tot=1;
 84     scanf("%d",&n);
 85     le=strlen(ss);
 86     for(i=0;i<le;i++){
 87         insert(ss+i);
 88     }
 89 }
 90 
 91 void check(int st,int now)
 92 {
 93     //printf(" st=%d v=%d now=%d\n",st,p[st].v,now);
 94     int y;
 95     if(now>k) return;
 96     ccount++;
 97     vector<int>::iterator it;
 98     for(it=p[st].nt.begin();it!=p[st].nt.end();it++)
 99     {
100         y=*it;
101         //printf("  st=%d y=%d f=%c\n",st,y,f[ p[y].v ]);
102         if(f[ p[y].v ]=='1'){
103             check(y,now);
104         }
105         else{
106             check(y,now+1);
107         }
108     }
109 }
110 
111 void solve()
112 {
113     //printf("solve\n");
114     ccount=-1;
115     check(0,0);
116 }
117 
118 void out()
119 {
120     printf("%d\n",ccount);
121 }
122 
123 int main()
124 {
125     //freopen("data.in","r",stdin);
126     //freopen("data.out","w",stdout);
127     //scanf("%d",&T);
128     //for(int ccnt=1;ccnt<=T;ccnt++)
129     //while(T--)
130     //scanf("%d%d",&n,&m);
131     while(scanf("%s",ss)!=EOF)
132     {
133         ini();
134         solve();
135         out();
136     }
137     return 0;
138 }

 

转载于:https://www.cnblogs.com/njczy2010/p/4289951.html