HDU6194-string string string

最新推荐文章于 2021-06-07 17:31:49 发布

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----线段树

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本文介绍了一种使用后缀数组和线段树解决字符串中子串出现次数问题的方法。通过构建后缀数组并结合线段树进行区间查询，有效地解决了寻找特定出现次数的子串数量的问题。

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string string string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1464 Accepted Submission(s): 422

Problem 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)≤105 ).
It's guaranteed that

∑length(s)≤2∗106 .

Output

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

Sample Input

  
   2
2
abcabc
3
abcabcabcabc

Sample Output

6
9

Source

2017 ACM/ICPC Asia Regional Shenyang Online

题意：给你一个字符串，问这个字符串中有多少个子串出现了k次

解题思路：后缀数组+线段树

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int N = 200010;

struct Sa
{
    char s[N];
    int rk[2][N], sa[N], h[N], w[N], now, k, n;
    int rmq[N][20], lg[N], bel[N];
    int mi[N * 4];
    LL ans;

    bool GetS()
    {
        scanf("%d", &k);
        scanf("%s", s + 1);
        return true;
    }

    void getsa(int z, int &m)
    {
        int x = now, y = now ^= 1;
        for (int i = 1; i <= z; i++) rk[y][i] = n - i + 1;
        for (int i = 1, j = z; i <= n; i++)
            if (sa[i] > z) rk[y][++j] = sa[i] - z;
        for (int i = 1; i <= m; i++) w[i] = 0;
        for (int i = 1; i <= n; i++) w[rk[x][rk[y][i]]]++;
        for (int i = 1; i <= m; i++) w[i] += w[i - 1];
        for (int i = n; i >= 1; i--) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];
        for (int i = m = 1; i <= n; i++)
        {
            int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];
            rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;
        }
    }

    void getsa(int m)
    {
        n = strlen(s + 1);
        now = rk[1][0] = sa[0] = s[0] = 0;
        for (int i = 1; i <= m; i++) w[i] = 0;
        for (int i = 1; i <= n; i++) w[s[i]]++;
        for (int i = 1; i <= m; i++) rk[1][i] = rk[1][i - 1] + (bool)w[i];
        for (int i = 1; i <= m; i++) w[i] += w[i - 1];
        for (int i = 1; i <= n; i++) rk[0][i] = rk[1][s[i]];
        for (int i = 1; i <= n; i++) sa[w[s[i]]--] = i;
        rk[1][n + 1] = rk[0][n + 1] = 0;
        for (int x = 1, y = rk[1][m]; x <= n&&y <= n; x <<= 1) getsa(x, y);
        for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : j)
        {
            if (rk[now][i] == 1) continue;
            int k = n - max(sa[rk[now][i] - 1], i);
            while (j <= k&&s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;
        }
    }

    void getrmq()
    {
        h[n + 1] = h[1] = lg[1] = 0;
        for (int i = 2; i <= n; i++)
            rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;
        for (int i = 1; (1 << i) <= n; i++)
        {
            for (int j = 2; j <= n; j++)
            {
                if (j + (1 << i) > n + 1) break;
                rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
            }
        }
    }

    int lcp(int x, int y)
    {
        int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);
        return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
    }

    void build(int k, int l, int r)
    {
        if (l == r) { mi[k] = h[l]; return; }
        int mid = (l + r) >> 1;
        build(k << 1, l, mid), build(k << 1 | 1, mid + 1, r);
        mi[k] = min(mi[k << 1], mi[k << 1 | 1]);
    }

    int query(int k, int l, int r, int ll, int rr)
    {
        if (l >= ll&&r <= rr) return mi[k];
        int mid = (l + r) >> 1, ans = INF;
        if (ll <= mid) ans = min(ans, query(k << 1, l, mid, ll, rr));
        if (rr > mid) ans = min(ans, query(k << 1 | 1, mid + 1, r, ll, rr));
        return ans;
    }

    void work()
    {
        getsa(300);
        build(1, 1, n);
        ans = 0;
        for (int i = 1; i + k - 1 <= n; i++)
        {
            int x = (k == 1 ? n - sa[i] + 1 : query(1, 1, n, i + 1, i + k - 1));
            int y = 0;
            y = max(y, h[i]);
            if (i + k - 1 != n) y = max(y, h[k + i]);
            if (x >= y) ans += 1LL * (x - y);
        }
        printf("%lld\n", ans);
    }
}sa;

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        sa.GetS();
        sa.work();
    }
    return 0;
}