CF666E Forensic Examination [后缀自动机，线段树合并]

标签：str   合并   sign   print   问题   const   efi   c++11   bool   

题意:

给出一个串 \(S\)，再给出 \(n\) 个串 \(T_i\)， \(q\) 次询问 \(S[pl,pr]\) 在 $ T_{[l,r]}$哪个串出现次数最多。

solution:

不难想到我们找 \(S[pl,pr]\) 是可以记录 \(ed_{pr}\) 然后倍增上去找到这个区间所对应的 SAM 节点。

我们把 \(T_i\) 插入 SAM 里，并且对应节点搞上 \(i\)，然后合并就好了qwq。

SAM 某个子树部分都是包含他自己的串，所以线段树合并一下就变成了子树数颜色以及找到最多颜色的问题了。

code:

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
using namespace std;
#define pb emplace_back
int n;
const int maxn = 1e6 + 61;
char s[maxn], buf[maxn];

struct PII {
  int x, y;
  bool operator<(const PII o) const { return x == o.x ? y > o.y : x < o.x; }
};

int rt[maxn];
struct SegMentTree {
  int ls[maxn << 5], rs[maxn << 5], cnt = 0;
  PII mx[maxn << 5], zero;

  SegMentTree() { zero.x = zero.y = 0; }

  void ins(int& p, int l, int r, int x) {
    if (!p) p = ++cnt;
    if (l == r) {
      mx[p].x++, mx[p].y = l;
      return;
    }
    int mid = l + r >> 1;
    if (x <= mid)
      ins(ls[p], l, mid, x);
    else
      ins(rs[p], mid + 1, r, x);
    mx[p] = max(mx[ls[p]], mx[rs[p]]);
  }

  int merge(int x, int y, int l, int r) {
    if (!x || !y) return x | y;
    int qwq = ++ cnt;
    if (l == r) {
    	mx[qwq] = mx[x];
      mx[qwq].x += mx[y].x;
      return qwq;
    }
    int mid = l + r >> 1;
    ls[qwq] = merge(ls[x], ls[y], l, mid);
    rs[qwq] = merge(rs[x], rs[y], mid + 1, r);
    mx[qwq] = max(mx[ls[qwq]], mx[rs[qwq]]);
    return qwq;
  }

  PII qry(int p, int a, int b, int l, int r) {
    if (!p) return zero;
    if (a <= l && r <= b) return mx[p];
    int mid = l + r >> 1;
    PII ans = zero;
    if (a <= mid) ans = max(ans, qry(ls[p], a, b, l, mid));
    if (b > mid) ans = max(ans, qry(rs[p], a, b, mid + 1, r));
    return ans;
  }
} smt;

vector<int> g[maxn];
int ed[maxn], mxl[maxn];

struct SAM {
  int ch[maxn][26], fa[maxn], len[maxn];
  int las, cnt;
  SAM() { las = cnt = 1; }

  void ins(int c, int id) {
    int p = las, np = las = ++cnt;
    smt.ins(rt[np], 1, n, id);
    len[np] = len[p] + 1;
    for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np;
    if (!p) {
      fa[np] = 1;
    } else {
      int q = ch[p][c];
      if (len[q] == len[p] + 1) {
        fa[np] = q;
      } else {
        int nq = ++cnt;
        memcpy(ch[nq], ch[q], sizeof(ch[q]));
        fa[nq] = fa[q], fa[q] = fa[np] = nq, len[nq] = len[p] + 1;
        for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
      }
    }
  }

  void build() {
    for (int i = 2; i <= cnt; i++) g[fa[i]].pb(i);
  }

  void init() {
    int p = 1, l = 0, now = 0;
    char* cur = s;
    while (*cur) {
      ++now;
      int c = (*cur++) - ‘a‘;
      while (p && !ch[p][c]) p = fa[p], l = len[p];
      if (ch[p][c]) {
        p = ch[p][c], ++l;
      } else {
        p = 1, l = 0;
      }
      ed[now] = p, mxl[now] = l;
    }
  }
} sam;

int fa[maxn][18];
void dfs(int u) {
  for (int v : g[u]) {
    fa[v][0] = u;
    dfs(v);
    rt[u] = smt.merge(rt[u], rt[v], 1, n);
  }
}

PII qry(int l, int r, int pl, int pr) {
  if (pr - pl + 1 > mxl[pr]) return { l, 0 };
  int p = ed[pr];
  for (int i = 17; ~i; i--)
    if (fa[p][i] && sam.len[fa[p][i]] >= pr - pl + 1) p = fa[p][i];
  PII ans = smt.qry(rt[p], l, r, 1, n);
  if (!ans.x) ans.y = l;
  return { ans.y, ans.x };
}

signed main() {
  // code begin.
  scanf("%s", s), scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    sam.las = 1, scanf("%s", buf);
    char* cur = buf;
    while (*cur) sam.ins((*cur++) - ‘a‘, i);
  }
  sam.build(), sam.init(), dfs(1);
  for (int j = 1; j <= 17; j++)
    for (int i = 1; i <= sam.cnt; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  int _;
  scanf("%d", &_);
  while (_--) {
    int l, r, pl, pr;
    scanf("%d %d %d %d", &l, &r, &pl, &pr);
    PII ans = qry(l, r, pl, pr);
    printf("%d %d\n", ans.x, ans.y);
  }
  return 0;
  // code end.
}

CF666E Forensic Examination [后缀自动机，线段树合并]

标签：str   合并   sign   print   问题   const   efi   c++11   bool   

原文地址：https://www.cnblogs.com/Isaunoya/p/12549379.html