poj 1743(后缀数组)

题意：给出一个数字序列，然后问这个序列的变化幅度的不重叠的最长重复子串。

题解：先把这个序列前后相差值计算出来，得到真正要找的最长重复子串的序列。方法是先用DA得到sa数组，然后算出height[i]：名次相邻的两个后缀串的最长公共前缀。用二分枚举最长重复子串的长度x，判断函数：只要连续的heighti出现了大于等于x(重复长度)且相差超过x(不重叠)，那么这个x就是有效的，二分出满足条件的最大值就是解。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 20005;
int wa
, wb
, ws
, wv
, sa
;
int rank
, height
, s
, n;

int cmp(int* r, int a, int b, int l) {
return (r[a] == r[b]) && (r[a + l] == r[b + l]);
}

void DA(int *r, int *sa, int n, int m) {
int i, j, p, *x = wa, *y = wb, *t;

for (i = 0; i < m; i++) ws[i] = 0;
for (i = 0; i < n; i++) ws[x[i] = r[i]]++;
for (i = 1; i < m; i++) ws[i] += ws[i - 1];
for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;

for (j = 1, p = 1; p < n; j *= 2, m = p) {

for (p = 0, i = n - j; 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 < n; i++) wv[i] = x[y[i]];
for (i = 0; i < m; i++) ws[i] = 0;
for (i = 0; i < n; i++) ws[wv[i]]++;
for (i = 0; i < m; i++) ws[i] += ws[i - 1];
for (i = n - 1; i >= 0; i--) sa[--ws[wv[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++;
}
}

void calheight(int *r, int *sa, int n) {
int i, j, k = 0;
for (i = 1; 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++);
}

//sa[i]表示排第i的位置
//height[i]表示sa[i]与sa[i - 1]最长公共前缀

int judge(int x) {
int l = sa[0], r = sa[0];
for (int i = 0; i <= n; i++) {
if (height[i] < x) {
l = r = sa[i];
continue;
}
l = min(l, sa[i]);
r = max(r, sa[i]);
if (r - l > x) return 1;
}
return 0;
}

int main() {
while (scanf("%d", &n) == 1 && n) {
for (int i = 0; i < n; i++)
scanf("%d", &s[i]);
for (int i = 0; i < n - 1; i++)
s[i] = s[i + 1] - s[i] + 100;
s[--n] = 0;
DA(s, sa, n + 1, 200);
calheight(s, sa, n);
int l = 1, r = n + 1;
while (l < r) {
int mid = (l + r + 1) / 2;
if (judge(mid))
l = mid;
else
r = mid - 1;
}
if (l < 4) printf("0\n");
else printf("%d\n", l + 1);
}
return 0;
}