poj 2406(后缀数组)

题意：给出一个串，是由它的一个子串重复k次得到的，问k最大是多少。
题解：从小到大枚举长度i，如果长度i的子串刚好是重复了len/i次，应该满足len % i == 0和rank[0] - rank[i] == 1 和height[rank[0]] == len-i这些条件的，直接判断就可以了，第一次用da倍增超时了(1000000的数据忘了。。。)，换成dc3过了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)
using namespace std;
const int N = 1000005;
int wa[N], wb[N], ws[N], wv[N], sa[N * 3];
int rank[N * 3], height[N * 3], s[N], n;
char str[N];

int c0(int *r, int a, int b) {
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}

int c12(int k, int *r, int a, int b) {
    if (k == 2)
        return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);
    return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}

void sort(int *r, int *a, int *b, int n, int m) {
    for (int i = 0; i < n; i++) wv[i] = r[a[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--) b[--ws[wv[i]]] = a[i];
}

void dc3(int *r, int *sa, int n, int m) {
    int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
    r[n] = r[n + 1] = 0;
    for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i;
    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
        rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;
    if (p < tbc) dc3(rn, san, tbc, p);
    else for (i = 0; i < tbc; i++) san[rn[i]] = i;
    for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3;
    if (n % 3 == 1) wb[ta++] = n - 1;
    sort(r, wb, wa, ta, m);
    for (i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i;
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
        sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    for (; i < ta; p++) sa[p] = wa[i++];
    for (; j < tbc; p++) sa[p] = wb[j++];
}

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++);
}

int main() {
    while (scanf("%s", str) == 1 && str[0] != '.') {
        int len = strlen(str);
        for (int i = 0; i < len; i++)
            s[i] = str[i] - 'a' + 1;
        s[len] = 0;
        dc3(s, sa, len + 1, 105);
        calheight(s, sa, len);
        int flag = 0;
        for (int i = 1; i <= len; i++) {
            if (len % i == 0 && rank[0] == rank[i] + 1 && height[rank[0]] == len - i) {
                printf("%d\n", len / i);
                flag = 1;
                break;
            }
        }
        if (!flag)
            printf("1\n");
    }
    return 0;
}