后缀数组

Max Substring

Time limit:  1000 ms
Memory limit:  256 MB

You are given a string $S$. Find a string $T$ that has the most number of occurrences as a substring in $S$.

If the solution is not unique, you should find the one with maximum length. If the solution is still not unique, find the smallest lexicographical one.

Standard input

The first line contains string $S$.

Standard output

Print string $T$ on the first line.

Constraints and notes

• $S$ consists of lowercase letters of the English alphabet
• The length of $S$ is between $1$ and 10^510​5​​ 

Input	Output	Explanation
cabdab	ab	a, b and ab appear $2$ times, but ab is bigger. There're no other substrings that appear more than once.
cabcabc	c	c appears $3$ times while a, b, ab,ca, bc, abc and cab appear only $2$ times.
ababababab	ab	Note that we're interested in substrings(continuous) not subsequences. ab is the winner with $5$ appearances.

后缀数组模板题目，练一下模板。

code：

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn=1000000;
int rankk[maxn],sa[maxn],height[maxn],tmp[maxn],cnt[maxn];
char s[maxn];
void suff(int n,int m)
{
    int i,j,k;
    n++;
//    printf("%d %d\n",n,m);
    for(i=0; i<n+10; i++)
        rankk[i]=sa[i]=height[i]=tmp[i]=0;
    for(i=0; i<m; i++)
        cnt[i]=0;
    for(i=0; i<n; i++)cnt[rankk[i]=s[i]]++;
    for(i=1; i<m; i++)
        cnt[i]+=cnt[i-1];
    for(i=0; i<n; i++)
        sa[--cnt[rankk[i]]]=i;
    for(k=1; k<=n; k<<=1)
    {
        for(i=0; i<n; i++)
        {
            j=sa[i]-k;
            if(j<0)
                j+=n;
            tmp[cnt[rankk[j]]++]=j;
        }
        sa[tmp[cnt[0]=0]]=j=0;
        for(i=1; i<n; i++)
        {
            if(rankk[tmp[i]]!=rankk[tmp[i-1]]||rankk[tmp[i]+k]!=rankk[tmp[i-1]+k])
                cnt[++j]=i;
            sa[tmp[i]]=j;
        }
        memcpy(rankk,sa,n*sizeof(int));
        memcpy(sa,tmp,n*sizeof(int));
        if(j>=n-1)
            break;
    }
    for(j=rankk[height[i=k=0]=0]; i<n-1; i++,k++)
        while(~k&&s[i]!=s[sa[j-1]+k])
            height[j]=k--,j=rankk[sa[j]+1];
}
int main()
{
    int n;
    scanf("%s",s);
    n=strlen(s);
    suff(n,'z'+1);

    int changdu=inf,geshu=0;
    int changduu=0,geshuu=0,biao=-1;
    for(int i=0; i<=n; i++)
    {
//        printf("%d\n",height[i]);
        if(height[i])
        {
            geshu++;
            changdu=min(changdu,height[i]);
            if(geshu>geshuu)
            {
                geshuu=geshu;
                changduu=changdu;
                biao=sa[i];
            }
            else  if(geshu==geshuu)
            {
                if(changdu>changduu)
                {
                    changduu=changdu;
                    biao=sa[i];
                }
            }
        }
        else
        {
            geshu=0;
            changdu=inf;
        }
    }
if(biao==-1)
    printf("%s",s);
else
    for(int i=biao; i<min(n,biao+changduu); i++)
        printf("%c",s[i]);
    printf("\n");
}