[CF] B. Segment Occurrences - Educational Codeforces Round 48 (Rated for Div. 2)

https://codeforces.com/contest/1016/problem/B

题目大意

给出 $n$ 长小写字符串 $s$ 和 $m$ 长小写字符串 $t$。给出 $q$ 组询问,每组询问给出 $l_i, r_i$,问 $s[l_i, r_i]$ 子串中 $t$ 出现了几次。

$1 \le n, m \le 1000, 1 \le q \le 10^5$。

简要题解

只要把所有匹配位置的尾巴标记为 $1$,这样就相当于询问 $[l_i + m - 1, r_i]$ 中有多少个 $1$,于是 KMP 处理匹配,然后前缀和处理一下就好了。

注意这题 $nm$ 的暴力匹配应该也可以因为范围给的很小。

复杂度

$T$:$O(n + m + q)$

$S$:$O(n + m)$

代码实现

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

int io_=[](){ ios::sync_with_stdio(false); cin.tie(nullptr); return 0; }();

using LL = long long;
using ULL = unsigned long long;
using LD = long double;
using PII = pair<int, int>;
using VI = vector<int>;
using MII = map<int, int>;

template<typename T> void cmin(T &x,const T &y) { if(y<x) x=y; }
template<typename T> void cmax(T &x,const T &y) { if(x<y) x=y; }
template<typename T> bool ckmin(T &x,const T &y) { 
    return y<x ? (x=y, true) : false; }
template<typename T> bool ckmax(T &x,const T &y) { 
    return x<y ? (x=y, true) : false; }
template<typename T> void cmin(T &x,T &y,const T &z) {// x<=y<=z 
    if(z<x) { y=x; x=z; } else if(z<y) y=z; }
template<typename T> void cmax(T &x,T &y,const T &z) {// x>=y>=z
    if(x<z) { y=x; x=z; } else if(y<z) y=z; }

// mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
// mt19937_64 rnd_64(chrono::system_clock::now().time_since_epoch().count());
namespace KMP {
vector<char> to_vec(const string& s) {
  vector<char> ans;
  for (char c : s) ans.push_back(c);
  return ans;
}

template<typename T>
vector<int> get_nxt(const vector<T>& a) {
  int n = a.size();
  vector<int> nxt(n + 1, -1);
  for (int i = 0, j = -1; i < n; ) {
    while (j != -1 && a[j] != a[i]) j = nxt[j];
    nxt[++i] = ++j;
  }
  return nxt;
}
vector<int> get_nxt(const string& s) {
  return get_nxt(to_vec(s));
}

/*
  kmp
  ans[i] maxlen of the prefix of t matches suffix of s[i]
*/
template<typename T>
vector<int> kmp(const vector<T>& s, const vector<T>& t) {
  int n = s.size(), m = t.size();
  vector<int> nxt = get_nxt(t);
  vector<int> ans(n);
  for (int i = 0, j = 0; i < n; i++) {
    while (j != -1 && (j == m || s[i] != t[j])) j = nxt[j];
    ans[i] = ++j;
  }
  return ans;
}

/*
  kmp_cnt
  ans[i] maxlen of the prefix of t matches suffix of s[i]
*/
template<typename T>
int kmp_cnt(const vector<T>& s, const vector<T>& t, bool overlap = true) {
  int n = s.size(), m = t.size();
  vector<int> len = kmp(s, t);
  int ans = 0, pre = -1;
  for (int i = 0; i < n; i++) {
    if (len[i] == m && (overlap || i - pre >= m)) {
      ans++;
      pre = i;
    }
  }
  return ans;
}
}
using namespace KMP;
/*
---------1---------2---------3---------4---------5---------6---------7---------
1234567890123456789012345678901234567890123456789012345678901234567890123456789
*/

void solve() {
  int n, m, q; cin >> n >> m >> q;
  string s, t; cin >> s >> t;

  auto a = kmp(to_vec(s), to_vec(t));
  
  vector<int> sum{0};
  for (int i = 0; i < n; i++) {
    sum.push_back(sum[i] + (a[i] == m));
  }

  int l, r;
  for (int i = 0; i < q; i++) {
    cin >> l >> r;

    l += m - 1;
    if (l > r) {
      cout << "0\n";
    } else {
      cout << (sum[r] - sum[l - 1]) << '\n';
    }
  }
}

int main() {
  int t = 1; 
  // cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
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