后缀数组应用——公共子串的个数

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长度不小于 K 的公共子串的个数(POJ 3415)

给出两个字符串 S 和 T，求他们长度不小于给定K 的公共子串的个数

首先把两个字符串拼接在一起，中间用一个没有出现过的字符做分隔符。求出新字符串的SA， Height 数组，类似求重复子串的方法，按 K 把 Height 数组分组。分别对于S 的每一个后缀，求他前面有多少个 T 的后缀与其的公共前缀大于 K。但是这样直接做复杂度是O( n ^ 2 )，时间无法承受。而我们可以知道两个后缀的公共前缀是他们形成的区间内 Height 的最小值，所以对于每一个后缀求解过程中前面的 Height 是递减的。用单调栈维护一下。再对 T 的每一个后缀按照前面的方法再做一次。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAX_N = 200005;int n, K, la, lb, a[MAX_N], sa[MAX_N], r[MAX_N], h[MAX_N];int ws[MAX_N], wv[MAX_N], wa[MAX_N], wb[MAX_N];char s[MAX_N >> 1];int f[MAX_N], st[MAX_N], cnt[MAX_N], tot = 0;long long ans = 0, reg = 0;void da(int *a, int *sa, int n, int m){int *x = wa, *y = wb;for (int i = 0; i < m; i ++) ws[i] = 0;for (int i = 0; i < n; i ++) ws[x[i] = a[i]] ++;for (int i = 1; i < m; i ++) ws[i] += ws[i - 1];for (int i = n - 1; i >= 0; i --) sa[-- ws[x[i]]] = i;for (int k = 1; k <= n; k <<= 1){int p = 0;for (int i = n - k; i < n; i ++)  y[p ++] = i;for (int i = 0; i < n; i ++) if (sa[i] >= k) y[p ++] = sa[i] - k;for (int i = 0; i < n; i ++) wv[i] = x[y[i]];for (int i = 0; i < m; i ++) ws[i] = 0;for (int i = 0; i < n; i ++) ws[wv[i]] ++;for (int i = 1; i < m; i ++) ws[i] += ws[i - 1];for (int i = n - 1; i >= 0; i --) sa[-- ws[wv[i]]] = y[i];swap(x, y); p = 1; x[sa[0]] = 0;for (int i = 1; i < n; i ++) x[sa[i]] = (y[sa[i - 1]] == y[sa[i]]) && (y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p ++;if (p >= n) break; m = p;}}void calc(){for (int i = 1; i <= n; i ++) r[sa[i]] = i;int k = 0, j;for (int i = 0; i < n; h[r[i ++]] = k)for (k ? k -- : 0, j = sa[r[i] - 1]; a[i + k] == a[j + k]; k ++);}void init(){scanf("%s", s); la = strlen(s);for (int i = 0; i < la; i ++) a[i] = s[i];a[la] = 1;scanf("%s", s); lb = strlen(s);for (int i = 0; i < lb; i ++) a[i + la + 1] = s[i];n = la + lb + 1; a[n] = 0;da(a, sa, n + 1, 128); calc();}void doit(){for (int i = 2; i <= n; i ++) {f[i] = sa[i] < la;h[i] = (h[i] - K + 1) > 0 ? h[i] - K + 1 : 0;}//for (int i = 1; i <= n; i ++) printf("%d ", h[i]); printf("\n");ans = 0; st[0] = -1; h[n + 1] = 0;for (int l = 0; l <= 1; l ++){tot = reg = 0;for (int i = 2; i <= n; i ++) {if (f[i] ^ l != 0) ans += reg;st[++ tot] = h[i + 1]; cnt[tot] = f[i] == l;//printf("%d\n", st[tot]);reg += (long long)st[tot] * cnt[tot];//for (int j = 1; j <= tot; j ++) printf("%d ", st[j]); printf("\n");while (st[tot - 1] >= st[tot]) {reg -= (long long)(st[tot - 1] - st[tot]) * cnt[tot - 1];st[tot - 1] = st[tot]; cnt[tot - 1] += cnt[tot];tot --;}}}printf("%I64d\n", ans);}int main(){while (scanf("%d", &K) != EOF) {if (!K) break;init();doit();}return 0;}

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