BZOJ4310 跳蚤（后缀数组+二分答案）

　　注意到答案一定是原串的子串，于是考虑造出SA，二分答案是第几小的子串。第k小子串很容易在SA上求出。之后计算使他成为最大子串至少要在几个位置切割，对每个字典序比答案大的后缀，找到所有合法切割位置（求lcp即可），就转化成了选最少的点使每个区间都包含至少一个点的经典问题。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 100010
#define ll long long
int n,m,sa[N],sa2[N],rk[N<<],tmp[N<<],h[N],cnt[N],l[N];
char s[N];
ll tot=;
void make()
{
    int m=;
    for (int i=;i<=n;i++) cnt[rk[i]=s[i]]++,m=max(m,(int)s[i]);
    for (int i=;i<=m;i++) cnt[i]+=cnt[i-];
    for (int i=n;i>=;i--) sa[cnt[rk[i]]--]=i;
    for (int k=;k<=n;k<<=)
    {
        int p=;
        for (int i=n-k+;i<=n;i++) sa2[++p]=i;
        for (int i=;i<=n;i++) if (sa[i]>k) sa2[++p]=sa[i]-k;
        memset(cnt,,sizeof(cnt));
        for (int i=;i<=n;i++) cnt[rk[i]]++;
        for (int i=;i<=m;i++) cnt[i]+=cnt[i-];
        for (int i=n;i>=;i--) sa[cnt[rk[sa2[i]]]--]=sa2[i];
        memcpy(tmp,rk,sizeof(rk));
        p=rk[sa[]]=;
        for (int i=;i<=n;i++)
        {
            if (tmp[sa[i]]!=tmp[sa[i-]]||tmp[sa[i]+k]!=tmp[sa[i-]+k]) p++;
            rk[sa[i]]=p;
        }
        if (p==n) break;
        m=p;
    }
    for (int i=;i<=n;i++)
    {
        h[i]=max(h[i-]-,);
        while (s[i+h[i]]==s[sa[rk[i]-]+h[i]]) h[i]++;
    }
}
bool check(ll k)
{
    int x=;
    for (int i=;i<=n;i++)
    if (k<=n-sa[i]+-h[sa[i]]) {x=i;break;}
    else k-=n-sa[i]+-h[sa[i]];
    k+=h[sa[x]];memset(l,,sizeof(l));
    int lcp=k;if (k<n-sa[x]+) l[sa[x]+lcp-]=max(l[sa[x]+lcp-],sa[x]);
    for (int i=x+;i<=n;i++)
    {
        lcp=min(lcp,h[sa[i]]);
        if (!lcp) return ;
        l[sa[i]+lcp-]=max(l[sa[i]+lcp-],sa[i]);
    }
    int t=;
    for (int i=;i<=n;i++)
    if (l[i])
    {
        t++;if (t>m) return ;
        int x=i;while (x<n&&l[x+]<=i) x++;
        i=x;
    }
    return ;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4310.in","r",stdin);
    freopen("bzoj4310.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    m=read();
    scanf("%s",s+);n=strlen(s+);
    make();
    tot=1ll*n*(n+)>>;
    for (int i=;i<=n;i++) tot-=h[i];
    ll l=,r=tot,ans;
    while (l<=r)
    {
        ll mid=l+r>>;
        if (check(mid)) r=mid-,ans=mid;
        else l=mid+;
    }
    for (int i=;i<=n;i++)
    if (ans<=n-sa[i]+-h[sa[i]]) {for (int j=sa[i];ans+h[sa[i]]>;j++,ans--)printf("%c",s[j]);break;}
    else ans-=n-sa[i]+-h[sa[i]];
    return ;
}