poj 3261 后缀数组 Or KMP 可重叠的 k 次最长重复子串

///题意： 求一个串至少出现k次的最长重复子串的长度，可重叠
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#define MAXN 22222
using namespace std;
int r[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], tmp[MAXN],a[MAXN];;
int sa[MAXN]; //index range 1~n value range 0~n-1
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, *ws = tmp;
    ///对r中长度为1的子串进行基数排序
    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;
     ///对r中长度为j的子串进行基数排序
    for (j = 1, p = 1; p < n; j *= 2, 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 (swap(x, y), 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++;
    }
}
int rank[MAXN]; //index range 0~n-1 value range 1~n
int height[MAXN]; //index from 1   (height[1] = 0)
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);
    return;
}
 int check(int mid,int n,int k)
{
      for(int sum=1,i =0; i <= n; i++) {
         if(height[i] >= mid){
             sum++;
             if(sum>=k) return 1;
        }else sum=1;
      }
     return 0;
}
int main()
{
    int n,k;
    //freopen("//media/学习/ACM/input.txt","r",stdin);
    while(scanf("%d%d", &n,&k) != EOF)
    {
        for(int i = 0; i < n; i++) scanf("%d", &a[i]),r[i] =a[i];
        r[n] = 0;
        da(r, sa, n + 1, MAXN);
        calheight(r, sa, n);
        int low =0, high = n, ans = -1;
        while(low <= high)
        {
            int mid = (low + high) >> 1;
            if(check(mid, n,k))
            {
                low = mid + 1;
                ans = max(ans, mid);
            }
            else high = mid - 1;
        }
        printf("%d\n",ans);
    }
    return 0;
}

或者 利用KMP 的next性质 水过~~~~~~~~~~~

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
int next[1000100];
char s[1000100];
void get_next(int n)
{
		int i=0;
		next[i]=-1;
		int j=next[0];

		while(i<=n)
		{
				if(j==-1 || s[i]==s[j])
				{
						i++;
						j++;
						next[i]=j;
				}
				else
						j = next[j];
		}
}
int main()
{

		scanf("%s",s);
		while(true)
		{
				if(s[0]=='.')
						break;
			int len = strlen(s);
			
			get_next(len);		

			int t = next[len];
			if(len%(len-t)==0)
					printf("%d/n", len/(len-t));
			else
					printf("%d/n",1);


			memset(s,0,sizeof(0));
			memset(next,0,sizeof(0));
			scanf("%s",s);
		}
}