POJ3693:Maximum repetition substring(后缀数组+RMQ)

The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing a ‘#‘.

For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

容易想到的就是枚举长度为L，然后看长度为L的字符串最多连续出现几次。

长度为L的串重复出现，那么st[0],st[l],st[2*l]……st[k*l]中肯定有两个连续的出现在字符串中。不然肯定长度不超过2*L啊。那么我们就枚举连续的两个，然后从这两个字符前后匹配，看最多能匹配多远。

即以st[i*l],st[i*l+l]前后匹配，这里是通过查询suffix(i*l),suffix(i*l+l)的最长公共前缀

通过rank值能找到i*l,与i*l+l的排名，我们要查询的是这段区间的height的最小值，通过RMQ预处理

达到查询为0(1)的复杂度，设LCP长度为M, 则答案显然为M / L + 1, 但这只是以i*l和i*l+l为起点的情况, 不过有一点是可以确定的。如果目标子串包含了

i*l和i*l+l。那么 i*l一定是和i*l+l匹配的。因为目标串中p一定和p+l匹配。这样才能满足子串长度为l。先在要解决的就是起点不在这两个位置上怎么办了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 1000005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank[N],height[N],s[N],a[N];
char str[N],str1[N],str2[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
int cmp(int *r,int a,int b,int k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包含末尾添加的0
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++)  wsf[i]=0;
    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
    p=1;
    j=1;
    for(; 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++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[wv[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
        t=x;
        x=y;
        y=t;
        x[sa[0]]=0;
        for(p=1,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
    }
}
void getheight(int *r,int n)//n不保存最后的0
{
    int i,j,k=0;
    for(i=1; i<=n; i++)  rank[sa[i]]=i;
    for(i=0; i<n; i++)
    {
        if(k)
            k--;
        else
            k=0;
        j=sa[rank[i]-1];
        while(r[i+k]==r[j+k])
            k++;
        height[rank[i]]=k;
    }
}

int Log[N];
int best[20][N];

void setLog(int n)
{
    Log[0] = -1;
    for(int i=1; i<=n; i++)
    {
        Log[i]=(i&(i-1))?Log[i-1]:Log[i-1] + 1 ;
    }
}
void RMQ(int n) {//初始化RMQ
    for(int i = 1; i <= n ; i ++) best[0][i] = height[i];
    for(int i = 1; i <= Log[n] ; i ++) {
        int limit = n - (1<<i) + 1;
        for(int j = 1; j <= limit ; j ++) {
            best[i][j] = min(best[i-1][j] , best[i-1][j+(1<<i>>1)]);
        }
    }
}
int lcp(int a,int b) {//询问a,b后缀的最长公共前缀
    a = rank[a];    b = rank[b];
    if(a > b) swap(a,b);
    a ++;
    int t = Log[b - a + 1];
    return min(best[t][a] , best[t][b - (1<<t) + 1]);
}
int ans[N],len,cas = 1;
int main()
{
    int i,j,k,n,l;
    setLog(N-1);
    W(~scanf("%s",str))
    {
        if(str[0]=='#') break;
        n = strlen(str);
        UP(i,0,n-1)
        s[i] = str[i];
        s[n] = 0;
        getsa(s,sa,n+1,300);
        getheight(s,n);
        RMQ(n);
        int maxn = -1;
        len = 0;
        UP(l,1,n-1)
        {
            for(i=0;i+l<n;i+=l)
            {
                k = lcp(i,i+l);
                int m = k/l+1;
                int t = l-k%l;
                t = i-t;
                if(t>=0 && k%l)
                {
                    if(lcp(t,t+l)>=k) m++;
                }
                if(m>maxn)
                {
                    len = 0;
                    ans[len++] = l;
                    maxn = m;
                }
                else if(m == maxn)
                {
                    ans[len++] = l;
                }
            }
        }
        int start,flag = 0;
        UP(i,1,n)
        {
            if(flag)
            break;
            UP(j,0,len-1)
            {
                int tem = ans[j];
                if(lcp(sa[i],sa[i]+tem)>=(maxn-1)*tem)
                {
                    start = sa[i];
                    l = tem*maxn;
                    flag = 1;
                    break;
                }
            }
        }
        printf("Case %d: ",cas++);
        UP(i,0,l-1)
        printf("%c",str[start+i]);
        printf("\n");
    }
}