001:Aggressive cows（二分搜索）

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该博客围绕 C/C++ 编程解决奶牛 stalls 分配问题。给定 N 个 stalls 位置和 C 头奶牛，要使任意两头奶牛间最小距离最大。给出输入输出要求及样例，提示处理大数据用 scanf，问题来源为 USACO 2005 February Gold。

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总时间限制: 1000ms 内存限制: 65536kB
描述
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?
输入

Line 1: Two space-separated integers: N and C
Lines 2…N+1: Line i+1 contains an integer stall location, xi
输出
Line 1: One integer: the largest minimum distance
样例输入
5 3
1
2
8
4
9
样例输出
3
提示
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.
来源
USACO 2005 February Gold

从1到1000000000中的数字作为牛之间的距离进行二分枚举，找到值后继续进行查找，如果不能找到距离就让r向左，找得到的话l向右
直到找出最小距离的最大值

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#define dd double
#define ll long long
dd PI = acos(-1);
using namespace std;
const ll MAXN = 100005;
const ll INF = 1e9 + 5;
ll n, m;
ll arr[MAXN];

bool binary(ll interval) {
	ll count = 0;
	ll j = 0;
	for (ll i = 1; i < n; i++) {
		if (arr[i] - arr[j] >= interval) {
			count++;
			j = i;
		}
		if (count == m - 1) {
			return true;
		}
	}
	return false;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for (ll i = 0; i < n; i++) {
		cin >> arr[i];
	}
	sort(arr, arr + n);
	ll l = 0;
	ll r = INF;
	ll mid;
	ll fin_mid = -1;
	while (l <= r) {
		mid = l + (r - l) / 2;
		if (binary(mid)) {
			fin_mid = max(fin_mid, mid);
			l = mid + 1;
		}
		else {
			r = mid - 1;
		}
	}
	cout << fin_mid << endl;
}