bzoj4892 [TJOI2017]DNA

给定一个匹配串和一个模式串，求模式串有多少个连续子串能够修改不超过 \(3\) 个字符变成匹配串

\(len\leq10^5\)

hash

---

枚举子串左端点，hash 求 lcp 枚举断点，接着跳过断点，记作一次修改，最多修改 \(3\) 次。特判修改 \(3\) 次后剩余部分是否相等。

时间复杂度 \(O(n\log n)\)

代码

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;

const int maxn = 1e5 + 10;

int Tests;

int n, m;

char str[maxn], s[maxn];

ull pw[maxn], sum1[maxn], sum2[maxn];

ull getsum(ull* a, int l, int r) {

  return a[r] - a[l - 1] * pw[r - l + 1];

}

int query(int x, int y) {

  int l = 0, r = m - y + 1, mid;

  while (l < r) {

    mid = (l + r + 1) >> 1;

    getsum(sum1, x, x + mid - 1) == getsum(sum2, y, y + mid - 1) ? l = mid : r = mid - 1;

  }

  return r;

}

int check(int x) {

  int y = 1;

  for (int i = 0; i < 3; i++) {

    int tmp = query(x, y);

    x += tmp + 1, y += tmp + 1;

    if (y > m) return 1;

  }

  return getsum(sum1, x, x + m - y) == getsum(sum2, y, m);

}

void solve() {

  scanf("%s %s", str + 1, s + 1);

  n = strlen(str + 1), m = strlen(s + 1);

  for (int i = 1; i <= n; i++) {

    sum1[i] = sum1[i - 1] * 131 + str[i];

  }

  for (int i = 1; i <= m; i++) {

    sum2[i] = sum2[i - 1] * 131 + s[i];

  }

  int ans = 0;

  for (int i = 1; i <= n - m + 1; i++) {

    ans += check(i);

  }

  printf("%d\n", ans);

}

int main() {

  pw[0] = 1;

  for (int i = 1; i < 100001; i++) {

    pw[i] = pw[i - 1] * 131;

  }

  scanf("%d", &Tests);

  while (Tests--) {

    solve();

  }

  return 0;

}