POJ 3693 Maximum repetition substring（后缀数组神题)

POJ 3693 Maximum repetition substring

题目链接

题意：给定一个字符串，求出其子串中，重复次数最多的串，如果有相同的，输出字典序最小的

思路：枚举长度l，把字符串按l分段，这样对于长度为l的字符串，肯定会包含一个分段位置，这样一来就可以在每个分段位置，往后做一次lcp，求出最大匹配长度，然后如果匹配长度有剩余，看剩余多少，就往前多少位置再做一次lcp，如果匹配出来长度更长，匹配次数就加1，这样就可以枚举过程中保存下答案了

这样问题还有字典序的问题，这个完全可以利用sa数组的特性，从字典序最小往大枚举，直到出现一个符合的位置就输出结束

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int INF = 0x3f3f3f3f;const int MAXLEN = 200005;struct Suffix {    int s[MAXLEN];    int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;    int rank[MAXLEN], height[MAXLEN];    int best[MAXLEN][20];    int len;    char str[MAXLEN];    int ans[MAXLEN], an;    void build_sa(int m) {n++;int i, *x = t, *y = t2;for (i = 0; i < m; i++) c[i] = 0;for (i = 0; i < n; i++) c[x[i] = s[i]]++;for (i = 1; i < m; i++) c[i] += c[i - 1];for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;for (int k = 1; k <= n; k <<= 1) {    int p = 0;    for (i = n - k; i < n; i++) y[p++] = i;    for (i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;    for (i = 0; i < m; i++) c[i] = 0;    for (i = 0; i < n; i++) c[x[y[i]]]++;    for (i = 0; i < m; i++) c[i] += c[i - 1];    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];    swap(x, y);    p = 1; x[sa[0]] = 0;    for (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;}n--;    }    void getHeight() {int i, j, k = 0;for (i = 1; i <= n; i++) rank[sa[i]] = i;for (i = 0; i < n; i++) {    if (k) k--;    int j = sa[rank[i] - 1];    while (s[i + k] == s[j + k]) k++;    height[rank[i]] = k;}    }    void initRMQ() {for (int i = 0; i < n; i++) best[i][0] = height[i + 1];for (int j = 1; (1<<j) <= n; j++)    for (int i = 0; i + (1<<j) - 1 < n; i++)best[i][j] = min(best[i][j - 1], best[i + (1<<(j - 1))][j - 1]);    }    int lcp(int L, int R) {L = rank[L] - 1; R = rank[R] - 1;if (L > R) swap(L, R);L++;int k = 0;while ((1<<(k + 1)) <= R - L + 1) k++;return min(best[L][k], best[R - (1<<k) + 1][k]);    }    void init() {n = 0;len = strlen(str);for (int i = 0; i < len; i++)    s[n++] = str[i] - 'a' + 1;s[n] = 0;    }    void solve() {init();build_sa(27);getHeight();initRMQ();int Max = 0;for (int l = 1; l < n; l++) {    for (int i = 0; i + l < n; i += l) {int tmp = lcp(i, i + l);int ti = tmp / l + 1;int v = i - (l - tmp % l);if (v >= 0 && tmp % l && lcp(v, v + l) >= tmp)    ti++;if (ti > Max) {    an = 0;    ans[an++] = l;    Max = ti;}else if (ti == Max)    ans[an++] = l;    }}int ans_v, ans_l;for (int i = 1; i <= n; i++) {    int flag = 0;    for (int j = 0; j < an; j++) {int tmp = ans[j];if (lcp(sa[i], sa[i] + tmp) >= (Max - 1) * tmp) {    ans_v = sa[i];    ans_l = Max * tmp;    flag = 1;}    }    if (flag) break;}for (int i = 0; i < ans_l; i++)    printf("%c", str[ans_v + i]);printf("\n");    }} gao;int main() {    int cas = 0;    while(~scanf("%s", gao.str) && gao.str[0] != '#') {printf("Case %d: ", ++cas);gao.solve();    }    return 0;}