CF700E Cool Slogans——SAM+线段树合并

RemoteJudge
又是一道用线段树合并来维护(endpos)的题，还有一道见我的博客CF666E

思路

先把(SAM)建出来
如果两个相邻的串(s_i)和(s_{i+1})要满足(s_i)在(s_{i+1})中至少出现了两次，那么(s_i)显然是(s_{i+1})对应的结点在(parent tree)上的祖先，那么我们可以在(parent tree)上树形dp来得出答案
转移自顶向下进行，(s_i)在(s_{i+1})中至少出现了两次意味着(s_i)在(s_{i+1})的所有(endpos)位置都出现了两次，所以我们只需要知道(s_{i+1})的(endpos)中任意一个元素并结合线段树来判断能否从(s_i)向(s_{i+1})转移。我直接维护了一个(firstpos)代表(endpos)中的最小值
最后注意不能转移时需要把值继承过来
代码

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define IINF 0x3f3f3f3f3f3f3f3fLL
#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back

#define N 200000

int n;
char s[N+5];

struct SAM {
  int nxt[26][2*N+5], maxlen[2*N+5], link[2*N+5], firstpos[2*N+5], tot, lst;
  int sumv[100*N+5], ch[2][100*N+5], root[2*N+5], nid;
  vi G[2*N+5];
  int top[2*N+5];
  ll f[2*N+5], ans;
  void init() {
    tot = lst = 1;
    nid = 0;
  }
  void pushup(int o) {
    sumv[o] = sumv[ch[0][o]]+sumv[ch[1][o]];
  }
  void add(int &o, int l, int r, int x) {
    if(!o) o = ++nid;
    if(l == r) {
      sumv[o] = 1;
      return ;
    }
    int mid = (l+r)>>1;
    if(x <= mid) add(ch[0][o], l, mid, x);
    else add(ch[1][o], mid+1, r, x);
    pushup(o);
  }
  int merge(int o, int u, int l, int r) {
    if(!o || !u) return o|u;
    int v = ++nid;
    if(l == r) {
      sumv[v] = sumv[o]+sumv[u] ? 1 : 0;
      return v;
    }
    int mid = (l+r)>>1;
    ch[0][v] = merge(ch[0][o], ch[0][u], l, mid);
    ch[1][v] = merge(ch[1][o], ch[1][u], mid+1, r);
    pushup(v);
    return v;
  }
  int query(int o, int l, int r, int L, int R) {
    if(!o) return 0;
    if(L <= l && r <= R) return sumv[o];
    int ret = 0, mid = (l+r)>>1;
    if(L <= mid) ret += query(ch[0][o], l, mid, L, R);
    if(R > mid) ret += query(ch[1][o], mid+1, r, L, R);
    return ret;
  }
  void extend(int c, int pos) {
    int cur = ++tot;
    maxlen[cur] = maxlen[lst]+1;
    firstpos[cur] = pos;
    while(lst && !nxt[c][lst]) nxt[c][lst] = cur, lst = link[lst];
    if(!lst) link[cur] = 1;
    else {
      int p = lst, q = nxt[c][p];
      if(maxlen[q] == maxlen[p]+1) link[cur] = q;
      else {
        int clone = ++tot;
        maxlen[clone] = maxlen[p]+1;
        link[clone] = link[q], link[q] = link[cur] = clone;
        firstpos[clone] = firstpos[q];
        for(int i = 0; i < 26; ++i) nxt[i][clone] = nxt[i][q];
        while(p && nxt[c][p] == q) nxt[c][p] = clone, p = link[p];
      }
    }
    lst = cur;
  }
  void dfs(int u) {
    for(int i = 0, v; i < G[u].size(); ++i) {
      v = G[u][i];
      dfs(v);
      root[u] = merge(root[u], root[v], 1, n);
    }
  }
  void build() {
    init();
    for(int i = 1; i <= n; ++i) {
      add(root[tot+1], 1, n, i);
      extend(s[i]-'a', i);
    }
    for(int i = 2; i <= tot; ++i) G[link[i]].pb(i);
    dfs(1);
  }
  void dp(int u) { // top数组用来辅助转移
    for(int i = 0, v; i < G[u].size(); ++i) {
      v = G[u][i];
      if(u == 1) f[v] = 1, top[v] = v;
      else {
        if(query(root[top[u]], 1, n, firstpos[v]-maxlen[v]+maxlen[top[u]], firstpos[v]) >= 2) f[v] = f[u]+1, top[v] = v;
        else f[v] = f[u], top[v] = top[u];
      }
      ans = max(ans, f[v]);
      dp(v);
    }
  }
  ll getans() {
    dp(1);
    return ans;
  }
}sa;

int main() {
  scanf("%d", &n);
  scanf("%s", s+1);
  sa.build();
  printf("%lld
", sa.getans());
  return 0;
}