本文主要是介绍POJ 3693 Maximum repetition substring（后缀数组神题)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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;
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