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);
}