HDU 6194 string string string （SAM）

Description

Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.

Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.

 

Input

The first line contains an integer T (T≤100) implying the number of test cases.

For each test case, there are two lines:

the first line contains an integer k (k≥1) which is described above;

the second line contain a string s (length(s)≤10^5).

 

Output

For each test case, print the number of the important substrings in a line.

 

Sample Input

22abcabc3abcabcabcabc

 

Sample Output

69

 

题意

询问字符串 s 中恰好出现 k 次的子串有多少个。

 

思路

和 HDU 4641 一样的思路。

区间减法计算 [至少出现 k 次的结果] 减去 [至少出现 k+1 次的结果] 即为 [恰好出现 k 次的结果]

 

AC 代码

#include <bits/stdc++.h>using namespace std;typedef __int64 LL;const int maxn = 2e5 + 10;LL ans;int K;struct SAM{    int next[maxn][26];    int pre[maxn], step[maxn];    int last, id;    int num1[maxn], num2[maxn];    void init()    {        last = id = 0;        memset(next[0], -1, sizeof(next[0]));        pre[0] = -1;        step[0] = 0;    }    void insert(int c)    {        int p = last, np = ++id;        step[np] = step[p] + 1;        memset(next[np], -1, sizeof(next[np]));        num1[np] = num2[np] = 0;        while (p != -1 && next[p][c] == -1)            next[p][c] = np, p = pre[p];        if (p == -1)            pre[np] = 0;        else        {            int q = next[p][c];            if (step[q] != step[p] + 1)            {                int nq = ++id;                memcpy(next[nq], next[q], sizeof(next[q]));                num1[nq] = num1[q];                num2[nq] = num2[q];                step[nq] = step[p] + 1;                pre[nq] = pre[q];                pre[np] = pre[q] = nq;                while (p != -1 && next[p][c] == q)                    next[p][c] = nq, p = pre[p];            }            else                pre[np] = q;        }        last = np;        /* diy */        bool f1 = false, f2 = false;        while (np != -1 && (!f1 || !f2))        {            if (!f1 && num1[np] < K)            {                num1[np]++;                if (num1[np] == K)                {                    ans += step[np] - step[pre[np]];                    f1 = true;                }            }            if (!f2 && num2[np] < K + 1)            {                num2[np]++;                if (num2[np] == K + 1)                {                    ans -= step[np] - step[pre[np]];                    f2 = true;                }            }            np = pre[np];        }        /* end diy */    }} sam;char str[maxn];LL get_ans(){    sam.init();    ans = 0;    int len = strlen(str);    for (int i = 0; i < len; i++)        sam.insert(str[i] - 'a');    return ans;}template <class T>inline void scan_d(T &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9')        ;    while (c >= '0' && c <= '9')    {        ret = ret * 10 + (c - '0'), c = getchar();    }}template <class T>inline void print_d(T x){    if (x > 9)    {        print_d(x / 10);    }    putchar(x % 10 + '0');}int main(){    int T;    scan_d(T);    while (T--)    {        scan_d(K);        gets(str);        print_d(get_ans());        putchar('\n');    }    return 0;}