围栏拆除(Fenced In P) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    long long A, B;
    int n, m;
    if (!(cin >> A >> B >> n >> m)) return 0;
    vector<long long> a(n + 2), b(m + 2);
    a[0] = 0;
    a[n + 1] = A;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    b[0] = 0;
    b[m + 1] = B;
    for (int i = 1; i <= m; ++i) cin >> b[i];
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    vector<long long> dx(n + 1), dy(m + 1);
    for (int i = 0; i <= n; ++i) dx[i] = a[i + 1] - a[i];
    for (int i = 0; i <= m; ++i) dy[i] = b[i + 1] - b[i];
    sort(dx.begin(), dx.end());
    sort(dy.begin(), dy.end());
    long long ans = dx[0] * m + dy[0] * n;
    int i = 1, j = 1;
    while (i <= n && j <= m) {
        if (dx[i] <= dy[j]) {
            ans += dx[i] * (m - j + 1);
            ++i;
        } else {
            ans += dy[j] * (n - i + 1);
            ++j;
        }
    }
    cout << ans << "\n";
    return 0;
}