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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;
}

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

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原文地址：http://blog.csdn.net/accelerator_/article/details/38780475