Description

给定n个字符串，询问每个字符串有多少子串（不包括空串）是所有n个字符串中至少k个字符串的子串？

Input

第一行两个整数n，k。

接下来n行每行一个字符串。

Output

一行n个整数，第i个整数表示第i个字符串的答案。

Sample Input

3 1
abc
a
ab

Sample Output

6 1 3

HINT

对于 100% 的数据，1<=n，k<=10^5，所有字符串总长不超过10^5，字符串只包含小写字母。

思路

首先发现这东西是真的不好做。。。那就找找性质

先把所有子串的贡献拆分成每个后缀的前缀的贡献

然后考虑怎么算每个后缀的贡献

又因为对于后缀i，假设前j个前缀有贡献，那么对于后缀i+1，它的前j-1个前缀一定是有贡献的

是不是就想到了height数组的处理，很容易证明这个的复杂度是线性的

那么怎么考虑可不可以扩展呢？

因为我们要算这个串的出现次数

并且知道连接所有串的分隔符一定是不会被匹配的

所以我们可以直接二分出包含这个串的排名区间

那么就可以预处理出排名第i的字符串至少要到第j个排名的字符串才能包含k个不同的字符串

这个东西可以双指针做

一定要把预处理的指针数组初值设成INF

 #include<bits/stdc++.h>using namespace std;typedef pair<int, int> pi;
typedef long long ll;
const int N = 2e5 + 10;
const int LOG = 20;struct Suffix_Array {int s[N], n, m;int c[N], x[N], y[N];int height[N], sa[N], rank[N];int st[N][LOG], Log[N];ll sum[N]; void init(int len, char *c) {n = len, m = 0;for (int i = 1; i <= len; i++) {s[i] = c[i];m = max(m, s[i]);}}void radix_sort() {for (int i = 1; i <= m; i++) c[i] = 0;for (int i = 1; i <= n; i++) c[x[y[i]]]++;for (int i = 1; i <= m; i++) c[i] += c[i - 1];for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];}void buildsa() {for (int i = 1; i <= n; i++) x[i] = s[i], y[i] = i;radix_sort();int now;for (int k = 1; k <= n; k <<= 1) {now = 0;for (int i = n - k + 1; i <= n; i++) y[++now] = i;for (int i = 1; i <= n; i++) if (sa[i] > k) y[++now] = sa[i] - k;radix_sort();y[sa[1]] = now = 1;for (int i = 2; i <= n; i++) y[sa[i]] = (x[sa[i]] == x[sa[i - 1]] && x[sa[i] + k] == x[sa[i - 1] + k]) ? now : ++now;swap(x, y);if (now == n) break;m = now;}}void buildrank() {for (int i = 1; i <= n; i++) rank[sa[i]] = i;}void buildsum() {for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + n - sa[i] + 1 - height[i];}void buildheight() {for (int i = 1; i <= n; i++) if (rank[i] != 1) {int k = max(height[rank[i - 1]] - 1, 0); for (; s[i + k] == s[sa[rank[i] - 1] + k]; k++);height[rank[i]] = k;}}void buildst() {Log[1] = 0;for (int i = 2; i < N; i++) Log[i] = Log[i >> 1] + 1;for (int i = 1; i <= n; i++) st[i][0] = height[i];for (int j = 1; j < LOG; j++) {for (int i = 1; i + (1 << (j - 1)) <= n; i++) {st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);}}}int queryst(int l, int r) {if (l == r) return n - sa[l] + 1;if (l > r) swap(l, r);++l; //***int k = Log[r - l + 1];return min(st[l][k], st[r - (1 << k) + 1][k]);}int querylcp(int la, int lb) {return queryst(rank[la], rank[lb]);}int querylcp(int la, int ra, int lb, int rb) {return min(min(ra - la + 1, rb - lb + 1), queryst(rank[la], rank[lb]));}void build(int len, char *c) {init(len, c);buildsa();buildrank();buildheight();buildsum();buildst();}
} Sa;char s[N], c[N];
int len[N], bg[N], ed[N], tot = 0;
int n, k, rpos[N], num[N], bel[N];bool check(int pos, int len) {int x = Sa.rank[pos], l, r;int l_line = x, r_line = x;l = 1, r = x - 1;while (l <= r) {int mid = (l + r) >> 1;if (Sa.queryst(x, mid) > len) l_line = mid, r = mid - 1;else l = mid + 1;} l = x + 1, r = tot;while (l <= r) {int mid = (l + r) >> 1;if (Sa.queryst(x, mid) > len) r_line = mid, l = mid + 1;else r = mid - 1;}return rpos[l_line] <= r_line;
}int main() {
#ifdef dream_makerfreopen("input.txt", "r", stdin);
#endifscanf("%d %d", &n, &k);for (int i = 1; i <= n; i++) {scanf("%s", c + 1);len[i] = strlen(c + 1);bg[i] = tot + 1;for (int j = 1; j <= len[i]; j++) {s[++tot] = c[j];bel[tot] = i;}ed[i] = tot;s[++tot] = '#';}Sa.build(tot, s);memset(rpos, 0x3f, sizeof(rpos)); //*****int l = 1, r = 0, cnt = 0;for (; r <= tot; r++) {if (!bel[Sa.sa[r]]) continue; ++num[bel[Sa.sa[r]]];if (num[bel[Sa.sa[r]]] == 1) ++cnt;if (cnt >= k) {for (; l <= r; l++) {if (!bel[Sa.sa[l]]) continue;if (cnt >= k) rpos[l] = r;else break;if (num[bel[Sa.sa[l]]] == 1) --cnt;--num[bel[Sa.sa[l]]];}} }for (int i = 1; i <= n; i++) {ll ans = 0; int cur = 0;for (int j = bg[i]; j <= ed[i]; j++) {cur = max(cur - 1, 0);while (j + cur <= ed[i] && check(j, cur)) ++cur;ans += cur;}printf("%lld ", ans);}return 0;
}