CF 900E Maximum Questions

题目描述

https://codeforces.com/problemset/problem/900/E

Solution

注意到在预处理之后我们可以 $O(1)$ 判断一段区间是否可以组成对应的子串

所以我们令 $f[i]$ 表示到 $i$ 为止，最多有多少个给定串，$g[i]$ 表示在 $f[i]$ 最大的情况下所需要的最少操作

时间复杂度 $O(n)$

#include <iostream>
#include <cstdio>
#define maxn 100010
using namespace std;

int n, m;

char s[maxn];

int a[maxn], b[maxn], p[maxn];

int f[maxn], g[maxn];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> s + 1 >> m; int la = m + 1 >> 1, lb = m / 2; 
    for (int i = 1; i <= n; ++i) 
        if (s[i] == 'a') a[i] = a[i - 2] + 1, b[i] = 0, p[i] = p[i - 1];
        else if (s[i] == 'b') b[i] = b[i - 2] + 1, a[i] = 0, p[i] = p[i - 1]; 
        else a[i] = a[i - 2] + 1, b[i] = b[i - 2] + 1, p[i] = p[i - 1] + 1; 
    for (int i = 1; i <= n; ++i) {
        f[i] = f[i - 1]; g[i] = g[i - 1];
        if (m & 1 && a[i] >= la && b[i - 1] >= lb || m % 2 == 0 && b[i] >= lb && a[i - 1] >= la)
            if (f[i - m] + 1 > f[i]) f[i] = f[i - m] + 1, g[i] = g[i - m] + p[i] - p[i - m];
            else if (f[i - m] + 1 == f[i]) g[i] = min(g[i], g[i - m] + p[i] - p[i - m]); 
    } cout << g[n] << "\n";
    return 0;
}