【SPOJ】LCS2

题面

https://vjudge.net/problem/SPOJ-LCS2

题解

#include<cstdio>
#include<iostream>
#include<cstring>
#define ri register int
#define N 100050
using namespace std;

char ss[N];
char s[N];
int ans[N<<1];

struct SAM {
  int ff[N<<1],len[N<<1];
  int ch[N<<1][26];
  int dp[N<<1];
  int ton[N<<1],c[N<<1];
  int tot,last;
  inline void clear() {
    tot=last=1;
  }
  inline void extend(int c) {
    int p=last,np=++tot; last=np;
    len[np]=len[p]+1;
    while (p && !ch[p][c]) ch[p][c]=np,p=ff[p];
    if (!p) ff[np]=1;
    else {
      int q=ch[p][c];
      if (len[q]==len[p]+1) {
        ff[np]=q;
      }
      else {
        int nq=++tot;
        for (ri i=0;i<26;i++) ch[nq][i]=ch[q][i]; ff[nq]=ff[q];
        len[nq]=len[p]+1;
        ff[np]=ff[q]=nq;
        while (p && ch[p][c]==q) ch[p][c]=nq,p=ff[p];
      }
    }
  }
  void getsa(){
    for (ri i=1;i<=tot;i++) ton[len[i]]++;
    for (ri i=1;i<=tot;i++) ton[i]+=ton[i-1];
    for (ri i=1;i<=tot;i++) c[ton[len[i]]--]=i;
  }
  void match() {
    memset(dp,0,sizeof(dp));
    int now=1,cnt=0;
    for (ri i=1,l=strlen(s+1);i<=l;i++) {
      if (ch[now][s[i]-'a']) {
        cnt++;
        now=ch[now][s[i]-'a'];
        if (cnt>dp[now]) dp[now]=cnt;
      }
      else {
        while (now && !ch[now][s[i]-'a']) now=ff[now];
        if (!now) {
          cnt=0;
          now=1;
          continue;
        }
        else {
          if (len[now]>dp[now]) dp[now]=len[now];
          cnt=len[now]+1;
          now=ch[now][s[i]-'a'];
          if (cnt>dp[now]) dp[now]=cnt;
        }
      }
    }
    for (ri i=tot;i>=1;i--) if (dp[c[i]]) dp[ff[c[i]]]=len[ff[c[i]]];
    for (ri i=1;i<=tot;i++) ans[i]=min(ans[i],dp[i]);
  }
} sam;

int main(){
  scanf("%s",ss+1);
  sam.clear();
  for (ri i=1,l=strlen(ss+1);i<=l;i++) sam.extend(ss[i]-'a');
  sam.getsa();
  memset(ans,0x3f,sizeof(ans));
  while (scanf("%s",s+1)==1) sam.match();
  int ret=0;
  for (ri i=1;i<=sam.tot;i++) if (ans[i]>ret) ret=ans[i];
  printf("%d
",ret);
}
原文地址:https://www.cnblogs.com/shxnb666/p/11279172.html