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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],b[N],sa[N],sa2[N],rk[N<<1],tmp[N<<1],cnt[N],h[N],f[N][20],g[N][20],lg2[N],ans[N],stk[N],top;
void make()
{
    int m=0;
    for (int i=1;i<=n;i++) cnt[rk[i]=a[i]]++,m=max(m,a[i]);
    for (int i=1;i<=m;i++) cnt[i]+=cnt[i-1];
    for (int i=n;i>=1;i--) sa[cnt[rk[i]]--]=i;
    for (int k=1;k<=n;k<<=1)
    {
        int p=0;
        for (int i=n-k+1;i<=n;i++) sa2[++p]=i;
        for (int i=1;i<=n;i++) if (sa[i]>k) sa2[++p]=sa[i]-k;
        memset(cnt,0,sizeof(cnt));
        for (int i=1;i<=n;i++) cnt[rk[i]]++;
        for (int i=1;i<=m;i++) cnt[i]+=cnt[i-1];
        for (int i=n;i>=1;i--) sa[cnt[rk[sa2[i]]]--]=sa2[i];
        memcpy(tmp,rk,sizeof(rk));
        p=rk[sa[1]]=1;
        for (int i=2;i<=n;i++)
        {
            if (tmp[sa[i]]!=tmp[sa[i-1]]||tmp[sa[i]+k]!=tmp[sa[i-1]+k]) p++;
            rk[sa[i]]=p;
        }
        if (p==n) break;
        m=p;
    }
    for (int i=1;i<=n;i++)
    {
        h[i]=max(h[i-1]-1,0);
        while (a[i+h[i]]==a[sa[rk[i]-1]+h[i]]) h[i]++;
    }
    for (int i=1;i<=n;i++) f[i][0]=h[sa[i]],g[i][0]=sa[i];
    lg2[1]=0;
    for (int i=2;i<=n;i++)
    {
        lg2[i]=lg2[i-1];
        if ((2<<lg2[i])<=i) lg2[i]++;
    }
    for (int j=1;j<20;j++)
        for (int i=1;i<=n;i++)
        f[i][j]=min(f[i][j-1],f[min(n,i+(1<<j-1))][j-1]),
        g[i][j]=min(g[i][j-1],g[min(n,i+(1<<j-1))][j-1]);
}
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+1]],f[y-(1<<lg2[y-x+1])+1][lg2[y-x+1]]);
}
int query2(int x,int y)
{
    return min(g[x][lg2[y-x+1]],g[y-(1<<lg2[y-x+1])+1][lg2[y-x+1]]);
}
int find(int x,int len)
{
    int l=1,r=x,u;
    while (l<=r)
    {
        int mid=l+r>>1;
        if (query(mid,x)>=len) u=mid,r=mid-1;
        else l=mid+1;
    }
    l=x,r=n;int v;
    while (l<=r)
    {
        int mid=l+r>>1;
        if (query(mid,x)>=len) v=mid,l=mid+1;
        else r=mid-1;
    }
    return query2(u,v);
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    m=read(),n=read();
    if (m==26) for (int i=1;i<=n;i++) a[i]=getc()-'a';
    else for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=n;i++) b[i]=a[i];
    sort(b+1,b+n+1);int t=unique(b+1,b+n+1)-b-1;
    for (int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+t+1,a[i])-b;
    make();
    int p=n+1;
    for (int i=1;i<=n;i++)
    {
        p=min(p,rk[i]);
        ans[n-i+1]=find(p,n-i+1);
    }
    for (int i=1;i<=n;i++) printf("%d ",ans[i]);
    return 0;
}