后缀数组（至少重复k次的可重叠的最长重复子串）—— POJ 3882

对应POJ 题目：点击打开链接

Stammering Aliens

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

Dr. Ellie Arroway has established contact with an extraterrestrial civilization. However, all efforts to decode their messages have failed so far because, as luck would have it, they have stumbled upon a race of stuttering aliens! Her team has found out that, in every long enough message, the most important words appear repeated a certain number of times as a sequence of consecutive characters, even in the middle of other words. Furthermore, sometimes they use contractions in an obscure manner. For example, if they need to say babtwice, they might just send the message babab, which has been abbreviated because the second b of the first word can be reused as the first b of the second one.

Thus, the message contains possibly overlapping repetitions of the same words over and over again. As a result, Ellie turns to you, S.R. Hadden, for help in identifying the gist of the message.

Given an integer m, and a string s, representing the message, your task is to find the longest substring of s that appears at least m times. For example, in the message baaaababababbababbab, the length-5 word babab is contained 3 times, namely at positions 5, 7 and 12(where indices start at zero). No substring appearing 3 or more times is longer (see the first example from the sample input). On the other hand, no substring appears 11 times or more (see example 2).

In case there are several solutions, the substring with the rightmost occurrence is preferred (see example 3).

Input

The input contains several test cases. Each test case consists of a line with an integer m ( m1), the minimum number of repetitions, followed by a line containing a string s of length between m and 40 000, inclusive. All characters in s are lowercase characters from ``a‘‘ to ``z‘‘. The last test case is denoted by m = 0 and must not be processed.

Output

Print one line of output for each test case. If there is no solution, output none; otherwise, print two integers in a line, separated by a space. The first integer denotes the maximum length of a substring appearing at least m times; the second integer gives the rightmost possible starting position of such a substring.

Sample Input

3
baaaababababbababbab
11
baaaababababbababbab
3
cccccc
0

Sample Output

5 12
none
4 2

思路：后缀数组的基础应用，二分答案分组求最大长度不是问题，反而在求下标那卡了一下。。。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MS(x, y) memset(x, y, sizeof(x))
const int MAXN = 40000+10;

int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN];
int rank[MAXN],r[MAXN],sa[MAXN],height[MAXN];
char str[MAXN];

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<<=1, 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=1; 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++;

	}
	return;
}

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-1; height[rank[i++]] = k)
		for(k ? k-- : 0,j=sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
	return;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	int n;
	while(~scanf("%d", &n), n)
	{
		MS(rank, 0);
		MS(sa, 0);
		MS(wa, 0);
		MS(wb, 0);
		MS(ws, 0);
		MS(wv, 0);
		MS(r, 0);
		MS(height, 0);
		scanf("%s", str);
		int len = strlen(str);
		if(1 == n){
			printf("%d 0\n", len);
			continue;
		}
		int maxn = 0;
		for(int i=0; i<len; i++){
			r[i] = str[i] - 'a' + 1;
			if(r[i] > maxn) maxn = r[i];
		}
		r[len++] = 0;//末尾添加一个最小值
		da(r, sa, len, maxn+1);
		calheight(r, sa, len);
#if 0
		printf("rank  : ");
		for(int i=0; i<len; i++)
			printf(" %d", rank[i]);
		printf("\n");

		printf("sa    : ");
		for(int i=0; i<len; i++)
			printf(" %d", sa[i]);
		printf("\n");

		printf("height: ");
		for(int i=0; i<len; i++)
			printf(" %d", height[i]);
		printf("\n");
#endif
		int left = 0, right = len-1;
		int mlen = 0, max1 = 0, max2 = 0, max3 = 0;
		int beg = 0, end = 0, ok;
		while(left <= right)
		{
			ok = 0;
			int mid = left + (right - left)/2;//二分答案
			max1 = max2 = 0;

			for(int i=2; i<len; i++){
				if(height[i] >= mid){//确定某一组的起点终点
					if(!beg) beg = i;
					end = i;
				}
				if((beg && end) && (i == len - 1 || height[i] < mid)){
					if(end - beg + 2 >= n){//符合题意的组
						max1 = 0;
						for(int i=beg-1; i<=end; i++)//求该组最右边的下标
							if(sa[i] > max1) max1 = sa[i];
						mlen = mid;
						if(max1 > max2) max2 = max1;
						ok = 1;
					}
					beg = end = 0;
				}
			}

			if(ok) max3 = max2;
			if(ok) left = mid + 1;
			else right = mid - 1;
		}
		if(mlen) printf("%d %d\n", mlen, max3);
		else printf("none\n");
	}
}