【每日一题】Problem 253B. Physics Practical

合集 - 每日一题(23)

1.【每日一题】Problem 363B. Fence2023-06-112.【每日一题】Problem 331C1. The Great Julya Calendar2023-06-103.【每日一题】Problem 327A - Flipping Game2023-06-094.【每日一题】Problem 313B - Ilya and Queries2023-06-085.【每日一题】Problem 1832B - Maximum Sum2023-06-076.【每日一题】Problem 44E. Anfisa the Monkey2023-06-127.【每日一题】Problem 120F. Spiders2023-06-138.【每日一题】Problem 174B. File List2023-06-149.【每日一题】Problem 180C. Letter2023-06-1510.【每日一题】Problem 189A. Cut Ribbon2023-06-1611.【每日一题】Problem 234C. Weather2023-06-22

12.【每日一题】Problem 253B. Physics Practical2023-06-22

13.【每日一题】Problem 443B. Kuriyama Mirai's Stones2023-06-2614.【每日一题】Problem 416B. Art Union2023-06-2715.【每日一题】Problem 489B. BerSU Ball2023-06-2716.【每日一题】Problem 476B. Dreamoon and WiFi2023-06-2917.【每日一题】Problem 489C. Given Length and Sum of Digits...2023-06-2918.【每日一题】Problem 505B. Mr. Kitayuta's Colorful Graph2023-07-0119.【每日一题】Problem 289B. Polo the Penguin and Matrix2023-07-0220.【每日一题】Problem 414B. Mashmokh and ACM2023-07-0221.【每日一题】Problem 522A. Reposts2023-07-0922.【每日一题】Problem 1195C. Basketball Exercise2023-07-1823.【每日一题】Problem 538B. Quasi Binary2023-07-19

收起

原题

解决思路

1. 定义 $dp[i][j]$ 为对 $i$ 元素做出选择后，要删除的最少元素个数
2. 对于 $i$，有两种情况，选或不选
   1. 选：则找到 $y(y>2x)$ 的个数，可以通过排序二分实现
   2. 不选：则在 $i-1$ 的最少删除个数的选择下 $+1$

#include <bits/stdc++.h>

int binarySearch(std::vector<int> &a, int target) {
    int l, r; l = 0, r = a.size() - 1;
    while (l <= r) {
        int mid = (r - l) / 2 + l;
        if (a[mid] <= target) l = mid + 1;
        else r = mid - 1;
    }
    return l;
}

int main() {
    std::freopen("input.txt", "r", stdin);
    std::freopen("output.txt", "w", stdout);

    int n;
    std::cin >> n;
    std::vector<int> c(n + 1, 0);
    for (int i = 1; i <= n; ++i) std::cin >> c[i];

    std::sort(c.begin() + 1, c.end());
    std::vector<std::vector<int>> dp(n + 1, std::vector<int>(2, 0));
    int index = binarySearch(c, c[1] * 2);
    dp[1][0] = 1;
    dp[1][1] = c.size() - index;
    for (int i = 2; i <= n; ++i) {
        index = binarySearch(c, c[i] * 2);
        dp[i][0] = std::min(dp[i - 1][0], dp[i - 1][1]) + 1;
        dp[i][1] = std::min(int(c.size()) - index + dp[i - 1][0], dp[i - 1][1]);
    }

    std::cout << std::min(dp[n][0], dp[n][1]) << std::endl;
}

更好的解

将二维数组替换为一维滚动数组

• 需要注意的是，$dp[0]$ 和 $dp[1]$ 都相互依赖对方更新前的值，因此需要先做一次备份

#include <bits/stdc++.h>

int binarySearch(std::vector<int> &a, int target) {
    int l, r; l = 0, r = a.size() - 1;
    while (l <= r) {
        int mid = (r - l) / 2 + l;
        if (a[mid] <= target) l = mid + 1;
        else r = mid - 1;
    }
    return l;
}

int main() {
    std::freopen("input.txt", "r", stdin);
    std::freopen("output.txt", "w", stdout);

    int n;
    std::cin >> n;
    std::vector<int> c(n + 1, 0);
    for (int i = 1; i <= n; ++i) std::cin >> c[i];

    std::sort(c.begin() + 1, c.end());
    std::vector<int> dp(2, 0);
    dp[1] = n + 1;
    for (int i = 1; i <= n; ++i) {
        int index = binarySearch(c, c[i] * 2);
        int t = dp[0]; // dp[1] 依赖更新前的 dp[0], 做一层备份
        dp[0] = std::min(dp[0], dp[1]) + 1;
        dp[1] = std::min(int(c.size()) - index + t, dp[1]);
    }

    std::cout << std::min(dp[0], dp[1]) << std::endl;
}