1021 字
5 分钟
后缀数组
后缀数组
详解 请参见 IOI2009 国家集训队论文
例3:不可重叠最长重复子串(pku1743)
先二分答案,判断是否存在两个长度为 k 的子串是相同的,且不重叠。
对于每组后缀,判断后缀的sa值的最大值和最小值之差是否不小于k
Code
#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int MAXN = 20010;int buc[MAXN], wa[MAXN], wb[MAXN], n;int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];void getheight(int n){ int i, j, k = 0; for (i = 0; i < n; i++) Rank[sa[i]] = i; for (i = 0; i < n; height[Rank[i++]] = k) for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++) ; return;}bool cmp(int *c, int a, int b, int d){ return c[a] == c[b] && c[a + d] == c[b + d];}void da(int n, int m = 320){ // for(int i=0;i<n;i++)printf("%d %d\n",i,r[i]); int i, j, p, *x = wa, *y = wb, *t; for (i = 0; i < m; i++) buc[i] = 0; for (i = 0; i < n; i++) buc[x[i] = r[i]]++; for (i = 1; i < m; i++) buc[i] += buc[i - 1]; for (i = n - 1; ~i; i--) sa[--buc[x[i]]] = i; for (j = 1, p = 1; p < n; j *= 2, m = p) { for (i = n - j, p = 0; 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++) buc[i] = 0; for (i = 0; i < n; i++) buc[x[y[i]]]++; for (i = 1; i < m; i++) buc[i] += buc[i - 1]; for (i = n - 1; ~i; i--) sa[--buc[x[y[i]]]] = y[i]; for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } // for(int i=0;i<n;i++) // printf("%d %d\n",i,sa[i]); getheight(n); return;}bool Judge(int mid){ int Min = 0x3f3f3f3f, Max = -0x3f3f3f3f; for (int i = 2; i <= n; i++) { if (height[i] < mid) Min = 0x3f3f3f3f, Max = -0x3f3f3f3f; else { Min = min(Min, min(sa[i - 1], sa[i])); Max = max(Max, max(sa[i - 1], sa[i])); if (Max - Min > mid) return 1; } } return 0;}int a[MAXN];int main(){ while (~scanf("%d", &n) && n) { for (int i = 0; i < n; i++) scanf("%d", &a[i]); n--; for (int i = 0; i < n; i++) r[i] = a[i + 1] - a[i] + 88; r[n] = 0; da(n + 1); int l = 0, r = (n >> 1) + 1, ans = 0; while (l < r) { int m = l + r >> 1; if(Judge(m)) ans = m, l = m + 1; else r = m; } if (ans >= 4) printf("%d\n", ans + 1); else printf("0\n"); }}
例4:可重叠的 k 次最长重复子串(pku3261)
先二分答案,然后将后缀分成若干组。判断有没有一个组的后缀个数不小于k。
Code
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 2000005;int buc[MAXN], wa[MAXN], wb[MAXN], n, K;int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];void GetHeight(int n){ int i, j, k = 0; for (i = 0; i < n; i++) Rank[sa[i]] = i; for (i = 0; i < n; height[Rank[i++]] = k) for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++) ; return;}bool cmp(int *c, int a, int b, int d){ return c[a] == c[b] && c[a + d] == c[b + d];}void da(int n, int m = 1000000){ int i, j, p, *x = wa, *y = wb, *t; for (i = 0; i < m; i++) buc[i] = 0; for (i = 0; i < n; i++) buc[x[i] = r[i]]++; for (i = 1; i < m; i++) buc[i] += buc[i - 1]; for (i = n - 1; ~i; i--) sa[--buc[x[i]]] = i; for (j = 1, p = 1; p < n; j *= 2, m = p) { for (i = n - j, p = 0; 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++) buc[i] = 0; for (i = 0; i < n; i++) buc[x[y[i]]]++; for (i = 1; i < m; i++) buc[i] += buc[i - 1]; for (i = n - 1; ~i; i--) sa[--buc[x[y[i]]]] = y[i]; for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } GetHeight(n); return;}bool Judge(int mid){ int l = 0; for (int i = 2; i <= n; i++) { if (height[i] < mid) l = 0; else { l++; if(l + 1 >= K) return 1; } } return 0;}int main(){ scanf("%d%d", &n, &K); for (int i = 0; i < n; i++) scanf("%d", &r[i]); da(n + 1); int l = 0, r = (n >> 1) + 1, ans = 0; while (l < r) { int mid = l + r >> 1; if (Judge(mid)) ans = mid, l = mid + 1; else r = mid; } printf("%d", ans);}