Poj 2456 Aggressive cows ( 二分+贪心

最新推荐文章于 2020-01-15 09:21:50 发布

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最新推荐文章于 2020-01-15 09:21:50 发布

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本文链接：https://blog.youkuaiyun.com/wang2332/article/details/70799402

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本文介绍了一个经典的算法问题：如何在给定的多个位置上放置一定数量的奶牛，使得任意两只奶牛之间的最小距离最大。文章通过二分查找和贪心算法解决了这一问题，并提供了完整的代码实现。

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Aggressive cows

Description

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

给N间房子，最大化最小值求每个奶牛的最大距离

题解：

二分求最大化最小值
先贪心将符合情况的d判断出来

先对房子排序

把第一头牛放进第一个房间内

将第i头牛扔进房子符合Xi+1 - Xi > d 寻找最大的Xi+1

然后二分查找所有情况

AC代码

#include <cstdio>
#include <algorithm>
using namespace std;
#define N 100010
int arr[N];
int n, m;
bool check(int d)   // check d的值 是否合理
{
    int f = 0;
    for(int i = 1;i < m; i++) {
        int cur = f+1;
        while(cur<n && arr[cur]-arr[f]<d) {
            cur++;
        }
        if(cur == n) return false;
        f = cur;
    }
return true;
 } 
int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 0;i < n; i++) 
        scanf("%d",&arr[i]);
    sort(arr,arr+n);
    int low = 0; int high = arr[n-1]-arr[0];
    while(high>=low) {
        int mid = (low+high) >> 1;
        if(check(mid)) low = mid+1;
        else high = mid-1;
    }
    printf("%d\n",low-1);
return 0;
}