使用后缀自动机求后缀数组

Posted 2020-10-17 z1j1n1

倒序建立后缀自动机的fail树就是后缀树，dfs后缀树得到后缀数组

#include <bits/stdc++.h>
using namespace std;

  int last,dis[200001],val[200001],cnt,a[200001][26],fa[200001],sa[200001],n,sta[200001],top;
  int tr[200001][26],rank[200001],height[200001];
  char st[200001];

  void ins(int num,int pos){
      int p=last,np=last=++cnt;
      dis[np]=dis[p]+1;val[np]=pos;
      while (!a[p][num]&&p) a[p][num]=np,p=fa[p];
      if (!p) fa[np]=1;else{
        int q=a[p][num];
        if (dis[p]+1==dis[q]) fa[np]=q;else{
          int nq=++cnt;dis[nq]=dis[p]+1;
          memcpy(a[nq],a[q],sizeof(a[q]));
          fa[nq]=fa[q];
          fa[q]=fa[np]=nq;
          while (a[p][num]==q) a[p][num]=nq,p=fa[p];
      }
    }
  }
  
  void dfs1(int po){
    for (int i=0;i<26;i++)
      if (a[po][i]&&dis[a[po][i]]==dis[po]+1){
        sta[++top]=i;
        tr[fa[a[po][i]]][sta[dis[a[po][i]]-dis[fa[a[po][i]]]]]=a[po][i];//确定后缀树的出边
        dfs1(a[po][i]);
        top--;
      }
  }
  
  void dfs2(int po){
    if (val[po]) sa[++cnt]=val[po];
    for (int i=0;i<26;i++)
      if (tr[po][i])
        dfs2(tr[po][i]);
  }
  
  void getheight(){
      int k=0;
      int po1,po2;
      for (int i=1;i<=n;i++){
          po1=i,po2=sa[rank[i]-1];
          if (k) k--;
          while ((st[po1+k]==st[po2+k])&&(po1+k<=n)&&(po2+k<=n))
          k++;
          height[rank[i]-1]=k;
      }
  }
  
  int main(){      
    scanf("%s",st+1);
    n=strlen(st+1);
    last=cnt=1;
    for (int i=n;i>=1;i--)
      ins(st[i]-‘a‘,i);
    
    dfs1(1);
    cnt=0;
    dfs2(1);
    for (int i=1;i<=n;i++) rank[sa[i]]=i;
    getheight();
    
    for (int i=1;i<=n;i++) printf("%d ",sa[i]);printf("\n");
    for (int i=1;i<n;i++) printf("%d ",height[i]);
  }