后缀数组模板

模板题：UOJ35
终于可以不看板子写出后缀数组了开心【捂脸】
但还是贴一下板子吧。
要记得开双倍数组qvq

#include <bits/stdc++.h>
#define N 200010
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define per(i,r,l) for (int i=r;i>=l;i--)
using namespace std;
int n,wa[N],wb[N],cnt[N],sa[N],r[N],h[N];
char s[N];

template <class Aqua>
inline void read(Aqua &s){
    s=0; Aqua f=1; char c=getchar();
    while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
    while (isdigit(c)) s=s*10+c-'0',c=getchar();
    s*=f;
}

inline bool cmp(int *y,int a,int b,int len){
    return (y[a]==y[b] && y[a+len]==y[b+len]);
}

void da(int m){
    int j,k,p(0),*x=wa,*y=wb;
    rep(i,0,m) cnt[i]=0;
    rep(i,1,n)
        cnt[x[i]=s[i]-96]++;
    rep(i,1,m)
        cnt[i]+=cnt[i-1];
    per(i,n,1)
        sa[cnt[x[i]]--]=i;
    for (j=p=1;p<n;j<<=1,m=p+1){
        k=0;
        rep(i,n-j+1,n)
            y[++k]=i;
        rep(i,1,n)
            if (sa[i]>j)
                y[++k]=sa[i]-j;
        rep(i,0,m) cnt[i]=0;
        rep(i,1,n)
            cnt[x[y[i]]]++;
        rep(i,1,m)
            cnt[i]+=cnt[i-1];
        per(i,n,1)
            sa[cnt[x[y[i]]]--]=y[i];
        for (swap(x,y),k=2,x[sa[p=1]]=1;k<=n;k++)
            x[sa[k]]=cmp(y,sa[k-1],sa[k],j)?p:++p;
    }
    rep(i,1,n)
        printf("%d ",sa[i]);
    puts("");
}

void getheight(){
    int ht=0;
    rep(i,1,n)
        r[sa[i]]=i;
    rep(i,1,n){
        if (ht) ht--;
        if (r[i]==1) continue;
        while (s[i+ht]==s[sa[r[i]-1]+ht]) ht++;
        h[r[i]]=ht;
    }
    rep(i,2,n)
        printf("%d ",h[i]);
    puts("");
}

int main(){
    scanf("%s",s+1);
    n=strlen(s+1);
    da(27);
    getheight();
    return 0;
}