后缀数组的应用——重复子串

重复子串：字符串 S
在字符串 T 中至少出现两次，则称 S 是 T 的重复子串

一、最长可覆盖的重复子串

只需求出字符串的 SA []，Height [] 数组，答案就是
Height 中最大的那个。

二、最长不可覆盖的重复子串 (POJ 1743)

求出 Height 数组，二分答案。问题就转变成了是否存在两个长度为 K 的相同字符串，且不重叠。按 K 分组，若
Height [ i ] < K 就重新分一组（如图）。对于每一组，如果这组中的 SA 的最大值与最小值的差大于等于 K 就存在不重叠的，否则不存在。

#include <cstdio>
#include <algorithm>

using namespace std;

const int MAX_N = 20005;

int wa[MAX_N], wb[MAX_N], ws[MAX_N], wv[MAX_N];
int n, a[MAX_N], sa[MAX_N], r[MAX_N], h[MAX_N];

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 = 0; 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 ++);
}

bool check(int x)
{
int mx = sa[0], mn = sa[0];
for (int i = 1; i < n; i ++){
if (h[i] < x) mx = mn = sa[i];
else {
if (sa[i] < mn) mn = sa[i];
if (sa[i] > mx) mx = sa[i];
if (mx - mn > x) return 1;
}
}
return 0;
}

void init()
{
int x = 0, y; scanf("%d", &x); n --;
for (int i = 0; i < n; i ++){
scanf("%d", &y); a[i] = y - x + 100;
x = y;
} a
= 0;
da(a, sa, n, 200); calc();
//	for (int i = 0; i < n; i ++) printf("%d ", sa[i]); printf("\n");
}

void doit()
{
int l = 1, r = n / 2;
while (l <= r){
int mid = (l + r) >> 1;
if (check(mid)) l = mid + 1;
else r = mid - 1;
}
if(r >= 4) printf("%d\n", r + 1);
else printf("0\n");
}

int main()
{
while (scanf("%d", &n) != EOF){
if (n == 0) break;
init();
doit();
}
return 0;
}

三、可覆盖 k 次的最长重复子串

和上一题的思路差不多，二分答案。判断时看每组的个数是否大于等于
k ，如果是就存在，否则不存在。

#include <cstdio>
#include <algorithm>

using namespace std;

const int MAX_N = 20005;

int n, K, a[MAX_N], sa[MAX_N], r[MAX_N], h[MAX_N];
int wa[MAX_N], wb[MAX_N], ws[1000005], wv[MAX_N];

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 = 0; 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 ++);
}
}

bool check(int x)
{
int cnt = 1;
for (int i = 1; i < n; i ++){
if (h[i] < x) cnt = 1;
else {
cnt ++;
if (cnt >= K) return 1;
}
}
return 0;
}

void init()
{
scanf("%d%d", &n, &K);
for (int i = 0; i < n; i ++)
scanf("%d", &a[i]);
da(a, sa, n, 1000000);
calc();
}

void doit()
{
int l = 1, r = n, mid, ans;
while (l <= r){
mid = (l + r) >> 1;
if (check(mid)) ans = mid, l = mid + 1;
else r = mid - 1;
}
printf("%d\n", ans);
}

int main()
{
init();
doit();
return 0;
}