Description

給定n個字符串，詢問每個字符串有多少子串（不包括空串）是所有n個字符串中至少k個字符串的子串？

Input

第一行兩個整數(shù)n，k。

接下來n行每行一個字符串。

Output

一行n個整數(shù)，第i個整數(shù)表示第i個字符串的答案。

Sample Input

3 1
abc
a
ab

Sample Output

6 1 3

HINT

對于 100% 的數(shù)據(jù)，1<=n，k<=10^5，所有字符串總長不超過10^5，字符串只包含小寫字母。

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思路

首先發(fā)現(xiàn)這東西是真的不好做。。。那就找找性質(zhì)

先把所有子串的貢獻(xiàn)拆分成每個后綴的前綴的貢獻(xiàn)

然后考慮怎么算每個后綴的貢獻(xiàn)

又因?yàn)閷τ诤缶Yi，假設(shè)前j個前綴有貢獻(xiàn)，那么對于后綴i+1，它的前j-1個前綴一定是有貢獻(xiàn)的

是不是就想到了height數(shù)組的處理，很容易證明這個的復(fù)雜度是線性的

那么怎么考慮可不可以擴(kuò)展呢？

因?yàn)槲覀円氵@個串的出現(xiàn)次數(shù)

并且知道連接所有串的分隔符一定是不會被匹配的

所以我們可以直接二分出包含這個串的排名區(qū)間

那么就可以預(yù)處理出排名第i的字符串至少要到第j個排名的字符串才能包含k個不同的字符串

這個東西可以雙指針做

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一定要把預(yù)處理的指針數(shù)組初值設(shè)成INF

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#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; }

轉(zhuǎn)載于:https://www.cnblogs.com/dream-maker-yk/p/10073765.html

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