Repeats SPOJ - REPEATS(后缀数组解决重复次数最多的连续重复子串)

标签：cal   char   公共前缀   +=   scanf   --   char s   poj   情况   

题目链接
题意：给定一个字符串，求重复次数最多的连续重复子串
题目思路：先穷举长度L，然后求长度为L的子串最多能连续出现几次。首先连续出现1次是肯定可以的，所以这里只考虑至少2次的情况。假设在原字符串中连续出现2次，记这个子字符串为S，那么S肯定包括了字符r[0], r[L], r[L2],r[L3], ……中的某相邻的两个。所以只须看字符r[Li]和r[L(i+1)]往前和
往后各能匹配到多远，记这个总长度为K，那么这里连续出现了K/L+1次。最后看最大值是多少。
穷举长度L的时间是n，每次计算的时间是n/L。所以整个做法的时间复杂度是O(n/1+n/2+n/3+……+n/n)=O(nlogn)。
用RMQ预处理区间最小值，方便求r[Li]和r[L(i+1)]的最长公共前缀，这就是往后面匹配了多长。
对于往前面匹配了多长，我们只要看jL-(i-k%i)和(j+1)L-(i-k%i)能否算上去即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
#define Max_N 50010
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) 
int wa1[Max_N*3], wb1[Max_N*3], wv1[Max_N*3], ws1[Max_N*3]; 
int c0(int *r, int a, int b)  
{
	return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
}
 
int c12(int k, int *r, int a, int b) 
{
	if(k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a+1, b+1); 
    else return r[a] < r[b] || r[a] == r[b] && wv1[a+1] < wv1[b+1];
}
    
 void sort(int *r, int *a, int *b, int n, int m) 
 { 
	int i; 
	for (i = 0; i < n; i++) wv1[i] = r[a[i]]; 
	for (i = 0; i < m; i++) ws1[i] = 0; 
	for (i = 0; i < n; i++) ws1[wv1[i]]++; 
	for (i = 1; i < m; i++) ws1[i] += ws1[i-1]; 
	for (i = n-1; i >= 0; i--) b[--ws1[wv1[i]]] = a[i]; 
	return; 
 }
void dc3(int *r, int *sa, int n, int m)
{
	int i, j, *rn = r + n, *san = sa + n,ta = 0,tb = (n + 1) / 3,tbc = 0,p; 
    r[n] = r[n+1] = 0; 
    for (i = 0; i < n; i++) if (i % 3 != 0) wa1[tbc++] = i; 
    sort(r+2, wa1, wb1, tbc, m); 
    sort(r+1, wb1, wa1, tbc, m); 
    sort(r, wa1, wb1, tbc, m); 
    for (p = 1,rn[F(wb1[0])] = 0,i = 1;i < tbc; i++) 
    rn[F(wb1[i])] = c0(r, wb1[i-1], wb1[i]) ? p - 1 : p++; 
    if (p < tbc) dc3(rn, san, tbc, p); 
    else for (i = 0; i < tbc; i++) san[rn[i]] = i;
    for (i = 0; i < tbc; i++) if (san[i] < tb) wb1[ta++] = san[i] * 3; 
    if (n % 3 == 1) wb1[ta++] = n - 1; 
    sort(r, wb1, wa1, ta, m); 
    for (i = 0; i < tbc; i++) wv1[wb1[i] = G(san[i])] = i; 
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++) 
    sa[p] = c12(wb1[j] % 3, r, wa1[i], wb1[j]) ? wa1[i++] : wb1[j++]; 
    for(; i < ta; p++) sa[p] = wa1[i++]; 
    for(; j < tbc; p++) sa[p] = wb1[j++]; 
    return; 
}
// str sa 都要开三倍的；
void da(int str[], int sa[], int rank1[], int height1[], int n, int m)
{
	for (int i = n; i < n*3; i++)
		str[i] = 0;
	dc3(str, sa, n+1, m);
	int i, j, k = 0;
	for (i = 0; i <= n; i++) rank1[sa[i]] = i;
	for (i = 0; i < n; i++)
	{
		if (k) k--;
		j = sa[rank1[i] - 1];
		while (str[i+k] == str[j+k]) k++;
		height1[rank1[i]] = k;
	}
	return;
}
int a[Max_N];
int str[3*Max_N];
int sa[3*Max_N];//sa[i]表示将所有后缀排序后第i小的后缀的编号。
int rank1[3*Max_N];//rank1[i]表示后缀i的排名。
int height1[3*Max_N]; //LCP(sa[i],sa[i-1]);
int n;
int minnum[Max_N][16];
void RMQ()   		//预处理  O(nlogn)
{
	int i,j;
	int m=(int)(log(n*1.0)/log(2.0));
	for(i=1;i<=n;i++)
		minnum[i][0]=height1[i];
	for(j=1;j<=m;j++)
		for(i=1;i+(1<<j)-1<=n;i++)
			minnum[i][j]=min(minnum[i][j-1],minnum[i+(1<<(j-1))][j-1]);
}
int Ask_MIN(int a,int b) 	//O(1)
{
	int k=int(log(b-a+1.0)/log(2.0));
	return min(minnum[a][k],minnum[b-(1<<k)+1][k]);
}
int calprefix(int a,int b)
{
    a=rank1[a],b=rank1[b];
    if(a>b)
        swap(a,b);
    return Ask_MIN(a+1,b);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		char s[5];
		for(int i=0;i<n;i++)
		{
			scanf("%s",s);
			str[i]=s[0]-‘a‘+1;
		}
		str[n]=0;
		da(str, sa, rank1, height1, n, 5);
		RMQ();
		int ans,Max=1;
		for(int i=1;i<=n;i++)
        {
            for(int j=0;j+i<n;j+=i)
            {
                ans=calprefix(j,j+i);
                int k=j-(i-ans%i);
                ans=ans/i+1;
                if(k>=0&&calprefix(k,k+i)>=i)
                    ans++;
                Max=max(Max,ans);
            }
        }
        printf("%d\n",Max);
	 } 
 }       

Repeats SPOJ - REPEATS(后缀数组解决重复次数最多的连续重复子串)

标签：cal   char   公共前缀   +=   scanf   --   char s   poj   情况   

原文地址：https://www.cnblogs.com/2462478392Lee/p/13810828.html