poj3693

题解：

后缀数组

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

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100005; 
int n,mx,ans,ansl,ansr,bin[30],Log[N],mn[N][17];
char ch[N];
void rmq(int mn[N][17],int *a)
{
    for (int i=1;i<=n;i++)mn[i][0]=a[i];
    for (int i=1;i<=Log[n];i++)
     for (int j=1;j<=n;j++)
      if (j+bin[i]-1<=n)mn[j][i]=min(mn[j][i-1],mn[j+bin[i-1]][i-1]);
      else break;
}
struct data
{
    int k,p,q,rk[2][N],sa[2][N],v[N],a[N],h[N],mn[N][17];    
    void ini()
     {
        memset(a,0,sizeof(a));
        memset(v,0,sizeof(v));
        memset(rk,0,sizeof(rk));
     }
    void cal(int *sa,int *rk,int *SA,int *RK)
     {
        for (int i=1;i<=n;i++)v[rk[sa[i]]]=i;
        for (int i=n;i;i--)
         if (sa[i]>k)SA[v[rk[sa[i]-k]]--]=sa[i]-k;
        for (int i=n-k+1;i<=n;i++)SA[v[rk[i]]--]=i;
        for (int i=1;i<=n;i++)
         RK[SA[i]]=RK[SA[i-1]]+(rk[SA[i]]!=rk[SA[i-1]]||rk[SA[i]+k]!=rk[SA[i-1]+k]);
     }
    void getsa()
     {
        p=0,q=1;
        for (int i=1;i<=n;i++)v[a[i]]++;
        for (int i=1;i<=30;i++)v[i]+=v[i-1];
        for (int i=1;i<=n;i++)sa[p][v[a[i]]--]=i;
        for (int i=1;i<=n;i++)
         rk[p][sa[p][i]]=rk[p][sa[p][i-1]]+(a[sa[p][i]]!=a[sa[p][i-1]]);
        for (k=1;k<n;k<<=1,swap(p,q))cal(sa[p],rk[p],sa[q],rk[q]);
     }
    void geth()
     {
        k=0;
        for (int i=1;i<=n;i++)
         if (rk[p][i]==1)h[rk[p][i]]=0;
         else 
          {
            int j=sa[p][rk[p][i]-1];
            while (a[i+k]==a[j+k])k++;
            h[rk[p][i]]=k;
            if (k>0)k--;
          }
     }
    void pre(){getsa();geth();rmq(mn,h);}
    int lcp(int x,int y)
     {
        x=rk[p][x];y=rk[p][y];
        if (x>y)swap(x,y);x++;
        int t=Log[y-x+1];
        return min(mn[x][t],mn[y-bin[t]+1][t]);
     } 
}c[2];
int query(int x,int y)
{
    int t=Log[y-x+1];
    return min(mn[x][t],mn[y-bin[t]+1][t]);
}
void solve(int L)
{
    int l=0,r=0,t;
    for (int i=1;i+L<=n;i+=L)
     if (ch[i]==ch[i+L])
      {
        r=c[0].lcp(i,i+L),l=c[1].lcp(n-i+2,n-i-L+2);
        if ((l+r)/L+1>mx)mx=(l+r)/L+1,ans=1e9;
        if ((l+r)/L+1==mx)
         {
            t=query(i-l,i-l+(l+r)%L);
            if (t<ans)
             {
                ans=t;
                ansl=c[0].sa[c[0].p][t];
                ansr=ansl+mx*L-1;
             }
         }
      }
}
int main()
{
    bin[0]=1;
    for (int i=1;i<20;i++)bin[i]=bin[i-1]<<1;
    Log[0]=-1;
    for (int i=1;i<=100000;i++)Log[i]=Log[i/2]+1;
    int tst=0;
    while (scanf("%s",ch+1))
     {
        if (ch[1]=='#')break;
        printf("Case %d: ",++tst);
        c[0].ini();
        c[1].ini();
        n=strlen(ch+1);
        for (int i=1;i<=n;i++)c[0].a[i]=ch[i]-'a'+1;
        for (int i=1;i<=n;i++)c[1].a[i]=ch[n-i+1]-'a'+1;
        c[0].pre();c[1].pre();
        rmq(mn,c[0].rk[c[0].p]);
        mx=1;ans=1e9;
        for (int i=1;i<=n;i++)
         if (c[0].rk[c[0].p][i]<ans)ans=c[0].rk[c[0].p][i],ansl=ansr=i;
        for (int i=1;i<=n;i++)solve(i);
        for (int i=ansl;i<=ansr;i++)putchar(ch[i]);
        puts("");
     }
    return 0;
}