Luogu 5108 仰望半月的夜空（后缀数组）

　　如果是要求左端点最大，直接求出SA，找前缀名次最小值就可以了。虽然现在要左端点最小，但我们已经知道了这个字典序最小的串是什么，找到名次数组上的合法区间求最小值即可。我也不知道为什么我会弃掉这个题，可能太久没写字符串了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 300010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
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;
}
int n,m,a[N],b[N],sa[N],sa2[N],rk[N<<],tmp[N<<],cnt[N],h[N],f[N][],g[N][],lg2[N],ans[N],stk[N],top;
void make()
{
    int m=;
    for (int i=;i<=n;i++) cnt[rk[i]=a[i]]++,m=max(m,a[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 (a[i+h[i]]==a[sa[rk[i]-]+h[i]]) h[i]++;
    }
    for (int i=;i<=n;i++) f[i][]=h[sa[i]],g[i][]=sa[i];
    lg2[]=;
    for (int i=;i<=n;i++)
    {
        lg2[i]=lg2[i-];
        if ((<<lg2[i])<=i) lg2[i]++;
    }
    for (int j=;j<;j++)
        for (int i=;i<=n;i++)
        f[i][j]=min(f[i][j-],f[min(n,i+(<<j-))][j-]),
        g[i][j]=min(g[i][j-],g[min(n,i+(<<j-))][j-]);
}
int query(int x,int y)
{
    if (x>y) swap(x,y);
    x++;if (x>y) return N;
    return min(f[x][lg2[y-x+]],f[y-(<<lg2[y-x+])+][lg2[y-x+]]);
}
int query2(int x,int y)
{
    return min(g[x][lg2[y-x+]],g[y-(<<lg2[y-x+])+][lg2[y-x+]]);
}
int find(int x,int len)
{
    int l=,r=x,u;
    while (l<=r)
    {
        int mid=l+r>>;
        if (query(mid,x)>=len) u=mid,r=mid-;
        else l=mid+;
    }
    l=x,r=n;int v;
    while (l<=r)
    {
        int mid=l+r>>;
        if (query(mid,x)>=len) v=mid,l=mid+;
        else r=mid-;
    }
    return query2(u,v);
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    m=read(),n=read();
    if (m==) for (int i=;i<=n;i++) a[i]=getc()-'a';
    else for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<=n;i++) b[i]=a[i];
    sort(b+,b+n+);int t=unique(b+,b+n+)-b-;
    for (int i=;i<=n;i++) a[i]=lower_bound(b+,b+t+,a[i])-b;
    make();
    int p=n+;
    for (int i=;i<=n;i++)
    {
        p=min(p,rk[i]);
        ans[n-i+]=find(p,n-i+);
    }
    for (int i=;i<=n;i++) printf("%d ",ans[i]);
    return ;
}