NC24017 [USACO 2016 Jan S]Angry Cows

题目

题目描述

Bessie the cow has designed what she thinks will be the next big hit video game: "Angry Cows". The premise, which she believes is completely original, is that the player shoots cows with a slingshot into a one-dimensional scene consisting of a set of hay bales located at various points on a number line. Each cow lands with sufficient force to detonate the hay bales in close proximity to her landing site. The goal is to use a set of cows to detonate all the hay bales.

There are N hay bales located at distinct integer positions x1,x2,…,xN on the number line. If a cow is launched with power R landing at position x, this will causes a blast of "radius R", destroying all hay bales within the range x−R…x+R.

A total of K cows are available to shoot, each with the same power R. Please determine the minimum integer value of R such that it is possible to use the K cows to detonate every single hay bale in the scene.

输入描述

The first line of input contains N (1≤N≤50,000) and K (1≤K≤10). The remaining N lines all contain integers x1…xN (each in the range 0…1,000,000,000).

输出描述

Please output the minimum power R with which each cow must be launched in order to detonate all the hay bales.

示例1

输入

7 2
20
25
18
8
10
3
1

输出

5

题解

思路

知识点：二分。

显然二分半径，由于覆盖范围是 \([x-R,x+R]\) 从 \(x-R\) 开始到另一点 \(x+R\) 一共距离 \(2R\) ，所以从之前一个点 \(a_{pre}\) 开始，到某个点 \(a_{pos}\) 的距离满足 \(a_{pos} - a_{pre}>2R\) 时，则让 \(a_{pos}\) 成为下一个 \(a_{pre}\) ，然后区域数 \(cnt\) 加一。如果最后 \(cnt \leq k\) ，则说明可行；\(cnt>k\)，则说明不可行。

时间复杂度 \(O(n \log \lceil \frac{a[n-1]-a[0]}{2} \rceil)\)

空间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>

using namespace std;

int a[50007];
int n, k;

bool check(int mid) {
    int pre = a[0], cnt = 0;
    for (int i = 1;i < n;i++) {
        if (a[i] - pre > 2 * mid) {
            cnt++;
            pre = a[i];
        }
    }
    cnt++;
    return cnt <= k;
}

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> k;
    for (int i = 0;i < n;i++) cin >> a[i];
    sort(a, a + n);
    int l = 1, r = (a[n - 1] - a[0] + 1) / 2;
    while (l <= r) {
        int mid = l + r >> 1;
        if (check(mid)) r = mid - 1;
        else l = mid + 1;
    }
    cout << l << '\n';
    return 0;
}

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来源： https://www.cnblogs.com/BlankYang/p/16421668.html