HDU 4641 K-string

K-string

Time Limit: 2000ms

Memory Limit: 131072KB

This problem will be judged on HDU. Original ID: 4641
64-bit integer IO format: %I64d      Java class name: Main

 

Given a string S. K-string is the sub-string of S and it appear in the S at least K times.It means there are at least K different pairs (i,j) so that Si,Si+1... Sj equal to this K-string. Given m operator or query:1.add a letter to the end of S; 2.query how many different K-string currently.For each query ,count the number of different K-string currently.

 

Input

The input consists of multiple test cases. 
Each test case begins with a line containing three integers n, m and K(1<=n,K<=50000,1<=m<=200000), denoting the length of string S, the number of operator or question and the least number of occurrences of K-string in the S.
The second line consists string S,which only contains lowercase letters. 
The next m lines describe the operator or query.The description of the operator looks as two space-separated integers t c (t = 1; c is lowercase letter).The description of the query looks as one integer t (t = 2).

 

Output

For each query print an integer — the number of different K-string currently.

 

Sample Input

3 5 2
abc
2
1 a
2
1 a
2

Sample Output

0
1
1

Source

2013 Multi-University Training Contest 4

 

解题：后缀自动机

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 600010;
int n,m,k;
LL ret;
struct node{
    int son[26],f,len,num;
    void init(){
        f = -1;
        num = len = 0;
        memset(son,-1,sizeof son);
    }
};
struct SAM{
    node e[maxn];
    int tot,last;
    int newnode(int len = 0){
        e[tot].init();
        e[tot].len = len;
        return tot++;
    }
    void init(){
        tot = last = 0;
        newnode();
    }
    void add(int c){
        int p = last,np = newnode(e[p].len + 1);
        while(p != -1 && e[p].son[c] == -1){
            e[p].son[c] = np;
            p = e[p].f;
        }
        if(p == -1) e[np].f = 0;
        else{
            int q = e[p].son[c];
            if(e[p].len + 1 == e[q].len) e[np].f = q;
            else{
                int nq = newnode();
                e[nq] = e[q];
                e[nq].len = e[p].len + 1;
                e[q].f = e[np].f = nq;
                while(p != -1 && e[p].son[c] == q){
                    e[p].son[c] = nq;
                    p = e[p].f;
                }
            }
        }
        for(p = np; p != -1; p = e[p].f){
            if(e[p].num == k) break;
            if(++e[p].num == k){
                ret += e[p].len - e[e[p].f].len;
                break;
            }
        }
        last = np;
    }
}sam;
char str[maxn];
int main(){
    int op;
    while(~scanf("%d%d%d",&n,&m,&k)){
        scanf("%s",str);
        sam.init();
        ret = 0;
        for(int i = 0; str[i]; ++i)
            sam.add(str[i] - 'a');
        for(int i = 0; i < m; ++i){
            scanf("%d",&op);
            if(op == 1){
                scanf("%s",str);
                sam.add(str[0] - 'a');
            }else printf("%d
",ret);
        }
    }
    return 0;
}