# 程序设计与算法（二）算法基础》《第四周 二分》Aggressive cows 2456

## 2456:Aggressive cows

1000ms

65536kB

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?

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

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

* Line 1: One integer: the largest minimum distance

5 3
1
2
8
4
9

3

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.

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

int a[100006];
int N, C;
bool func(int x)
{
int temp = a[0];
int cnt = 1;
for (int i = 1; i < N; i++)
{
//统计在设定间距下（比如4），牛舍的位置是否放的下C头牛
if (a[i] - temp >= x)
{
cnt++;
temp = a[i];
}
if (cnt == C)
{
return true;
}
}
return false;
}

int main()
{
int i;
cin >> N; // room number for cows
cin >> C; // cows number
for (i = 0; i < N; i++)
{
scanf("%d", &a[i]);
}
sort(a, a + N);

int  left, right, mid;
left = 0;
right = a[N - 1]-a[0];
//mid = left + (right - left) / 2;
while (left <= right)
{// 5个牛舍的位置【1， 2， 4， 8， 9】，最大间距是 8，所有可能间距取值是【0~8】，二分查找问题
mid = left + (right - left) / 2;//第一把间距是4，失败，需要缩小间距
if ((func(mid)))  // 如果现在的间距满足放下C头牛，那要试一试间距变大的情况
{ // 间距如何变大？
left = mid + 1; /* left = 5 ,[0~4]delete,  [5~8]select*/
}
else
{// 间距缩小
right = mid - 1; /*right = 3, [3~8]delete,  [0~3]select */
}
}
// 当间距=3时，满足成立条件，return true, 此时，left=3+1=4, while(4<=3)退出，实际间距4-1=3
printf("%d\n",left-1);

return 0;
}

posted @ 2019-09-15 13:14  清风oo  阅读(227)  评论(0编辑  收藏  举报