二分枚举模板 poj2456 疯牛

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于 2019-06-11 13:22:21 发布

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分类专栏：二分文章标签：二分

本文链接：https://blog.youkuaiyun.com/qq_33914105/article/details/91427997

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探讨了在给定的N个位置中选择C头牛，使得任意两头牛之间的最小距离最大化的算法。通过二分查找和贪心策略，找到满足条件的最大最小距离。

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Aggressive cows

POJ - 2456

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

Input

* Line 1: Two space-separated integers: N and C

* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

* Line 1: One integer: the largest minimum distance

Sample Input

Sample Output

Hint

OUTPUT DETAILS:

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.

Huge input data,scanf is recommended.

题意：简单的说就是给你一段长度，在这一段中给出m个点，然后在这m个点中选出k个点，让这k个点之间相邻两个点的之间距离的最小值最大

思路：通过二分枚举这个最小值，然后通过贪心的思想找出满足要求的最大的这个最小值

解析：——二分枚举 + 贪心

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int maxn = 1e5 + 5;
int m[maxn];
int n, c;
bool property(int q){ //枚举
	int cnt = 1;
    int st = m[0];
    for(int i = 1; i < n; i++)
    {	
    			
        if(m[i] - st >= q)
        {
            ++cnt;
            if(cnt >= c)
                return true; 
            st = m[i];
        }
    }
    return false;
} 
int binary_search(){ //二分
	int l = 0;
	int r = m[n - 1] - m[0];
	int mid = (l + r) >> 1;
	while(l <= r){
		if(property(mid))
			l = mid + 1;
		else 
			r = mid - 1;
		mid = (l + r) >> 1; 
	}
	return l - 1;
}
int main(){
	cin >> n >> c;
	for(int i = 0; i < n; i++){
		scanf("%d",&m[i]);
	}
	sort(m, m + n);
	cout << binary_search() << endl;
}