hdu-6194 string string string 后缀数组 出现恰好K次的串的数量

Posted 2020-10-07 Luke_Ye

 最少出现K次我们可以用Height数组的lcp来得出，而恰好出现K次，我们只要除去最少出现K+1次的lcp即可。

#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <iostream>
using namespace std;

const int maxn = 100000 + 10;

int t1[maxn], t2[maxn], c[maxn];

bool cmp(int* r, int a, int b, int l) {
    return r[a] == r[b] && r[a + l] == r[b + l];
}

void da(int str[], int sa[], int Rank[], int lcp[], int n, int m) {
    ++n;
    int i, j, p, *x = t1, *y = t2;
    for (i = 0; i < m; ++i) c[i] = 0;
    //        puts("hha");  
    for (i = 0; i < n; ++i) c[x[i] = str[i]]++;
    for (i = 1; i < m; ++i) c[i] += c[i - 1];
    for (i = n - 1; i >= 0; --i) sa[--c[x[i]]] = i;
    for (j = 1; j <= n; j <<= 1) {
        p = 0;
        for (i = n - j; i < n; ++i) y[p++] = i;
        for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
        for (i = 0; i < m; ++i) c[i] = 0;
        for (i = 0; i < n; ++i) c[x[y[i]]]++;

        for (i = 1; i < m; ++i) c[i] += c[i - 1];
        for (i = n - 1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];

        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for (i = 1; i < n; ++i) {
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
        }
        if (p >= n)break;
        m = p;
    }

    int k = 0;
    n--;
    for (i = 0; i <= n; ++i) Rank[sa[i]] = i;
    for (i = 0; i < n; ++i) {
        if (k)--k;
        j = sa[Rank[i] - 1];
        while (str[i + k] == str[j + k])++k;
        lcp[Rank[i]] = k;
        //cout << k << endl;
    }
}

int lcp[maxn], a[maxn], sa[maxn], Rank[maxn];

char s[maxn];

int d[maxn][40];
int len;

void rmq_init(int* A, int n) {
    for (int i = 0; i < n; ++i) d[i][0] = A[i];
    for (int j = 1; (1 << j) <= n; ++j)
        for (int i = 0; i + (1 << j) - 1 < n; ++i)
            d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}

int ASK(int l, int r) {
    int k = 0;
    while ((1 << (k + 1)) <= r - l + 1)++k;
    return min(d[l][k], d[r - (1 << k) + 1][k]);
}

int ask(int l, int r) {
    if (l == r) return len - sa[r]; /// l == r的话 是一个串， 返回本身的长度即可。  
    return ASK(l + 1, r); ///否则在rmq查询。  
}

//  
int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("outstd.txt", "w", stdout);
    int T;
    scanf("%d", &T);
    while (T--) {
        int k;
        scanf("%d", &k);
        scanf("%s", s);
        //puts(s);
        len = strlen(s);
        for (int i = 0; i < len; ++i) {
            a[i] = s[i] - ‘a‘ + 1;
        }
        a[len] = 0;
        da(a, sa, Rank, lcp, len, 30);
        rmq_init(lcp, len + 1);
        long long ans = 0;
        if (k == 1)
        {
            for (int i = 1; i <= len; i++)
            {
                int siz = len - sa[i];
                int des = 0;
                if (i > 1) des = max(des, ask(i - 1, i));
                if (i < len) des = max(des, ask(i, i + 1));
                ans += siz - des;
            }
        }
        else
        {
            for (int i = 1; i+k-1 <= len; i++)
            {
                int siz = ask(i, i + k - 1);
                int des = 0;
                if (i > 1) des = max(des, ask(i - 1, i + k - 1));
                if (i + k <= len) des = max(des, ask(i, i + k));
                ans += siz - des;
            }
        }
        printf("%I64d\n", ans);
        /*
        long long ans = 0;

        for (int i = 1; i + k - 1 <= len; ++i) {
            ans += ask(i, i + k - 1);
            if (i - 1 > 0)ans -= ask(i - 1, i + k - 1); ///注意边界问题。  
            if (i + k <= len)ans -= ask(i, i + k);
            if (i - 1 > 0 && i + k <= len)ans += ask(i - 1, i + k);
        }
        printf("%I64d\n", ans);*/


    }
    return 0;
}