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[N], wb[N], ws[N], wv[N], sa[N];int rank[N], height[N], s[N], 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;}