Aggressive cows

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 23866   Accepted: 11141

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

5 3
1
2
8
4
9

Sample Output

3

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.

 

尝试了剪枝 结果运行时间并没有什么区别(注释掉的地方是剪枝)

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
const int si = 100010, inf = 1000000000 + 10;
using namespace std;
int N, COW;
int ar[si]; bool c(int dis) {
int pre = 0, finished = 1;
for (int i = 1; i < N && finished < COW; i++) {
//if (ar[N - 1] - ar[pre] < dis * (COW - finished)) return false;
//剪枝 剩下COW - finished头牛 每头牛需要dis的距离
if (ar[i] - ar[pre] >= dis) {
finished++;
pre = i;
}
}
return finished >= COW;
}
int main() {
cin >> N >> COW;
for (int i = 0; i < N; i++) scanf("%d", &ar[i]);;
sort(ar, ar + N);
int l = 1, m, r = inf;
while (l < r - 1) {
m = l + r >> 1;
if (c(m)) l = m;
else r = m;
}
cout << l << endl;
return 0;
}

2456 Aggressive cows的更多相关文章

  1. poj 2456 Aggressive cows && nyoj 疯牛 最大化最小值 二分

    poj 2456 Aggressive cows && nyoj 疯牛 最大化最小值 二分 题目链接: nyoj : http://acm.nyist.net/JudgeOnline/ ...

  2. 二分搜索 POJ 2456 Aggressive cows

    题目传送门 /* 二分搜索:搜索安排最近牛的距离不小于d */ #include <cstdio> #include <algorithm> #include <cmat ...

  3. POJ 2456 Aggressive cows (二分 基础)

    Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7924   Accepted: 3959 D ...

  4. [ACM] poj 2456 Aggressive cows (二分查找)

    Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5436   Accepted: 2720 D ...

  5. POJ 2456 Aggressive cows

    Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11192   Accepted: 5492 ...

  6. POJ 2456 Aggressive cows(二分答案)

    Aggressive cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22674 Accepted: 10636 Des ...

  7. POJ - 2456 Aggressive cows 二分 最大化最小值

    Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18099   Accepted: 8619 ...

  8. poj 2456 Aggressive cows 贪心+二分

    Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25944   Accepted: 11982 ...

  9. POJ 2456 Aggressive cows(贪心 + 二分)

    原题链接:Aggressive cows 题目大意:农夫 建造了一座很长的畜栏,它包括  个隔间,这些小隔间依次编号为. 但是, 的  头牛们并不喜欢这种布局,而且几头牛放在一个隔间里,他们就要发生争 ...

随机推荐

  1. 4A Watermelon

    A. Watermelon time limit per test 1 second memory limit per test 64 megabytes input standard input o ...

  2. CNN - 卷积神经网络

    例:  卷积公式: 卷积和卷积没有什么特别的关系,只是计算步骤比较像,成为卷积神经网络名字的由来. 感受野:单个感觉神经元的感受野是感觉空间的特定区域(如体表或视野),在这个区域内,刺激会改变神经元的 ...

  3. 利用JS打印质数

    我爱撸码,撸码使我感到快乐!大家好,我是Counter,今天非常愉快,没有前几天的相对比较复杂的逻辑思维在里面,今天来写写,利用JS打印质数,基本上很多面试,会很经常的考到.那废话不多说,直接上代码: ...

  4. Neo4j 安装插件APOC和GRAPH ALGORITHMS

    在 https://github.com/neo4j-contrib/neo4j-apoc-procedures/releases 下载apoc扩展包JAR文件 在 https://github.co ...

  5. webpack 与 vue 打包体积优化

    webpack 与 vue 在使用vue开发时,遇到打包后单个文件太大,因而需要分包,不然加载时间太久.虽然尽可能减少请求次数,但是单个包太大也不是好事 思路 组件按需加载 vue-router 的懒 ...

  6. python学习-----协程

    一.协程的引入 对于单线程下,我们不可避免程序中出现io操作,但如果我们能在自己的程序中(即用户程序级别,而非操作系统级别)控制单线程下的多个任务能在一个任务遇到io阻塞时就切换到另外一个任务去计算, ...

  7. vue--一些预设属性

    Babel=> 转换工具,将ES6转换成ES5的转换工具.有些浏览器不支持ES6语法: router=> 路由管理器: vueX=> 一个仓库,存储状态信息: CSS pre-pro ...

  8. css js 兼容问题

    js  兼容问题 1. document.form.item 问题问题:代码中存在 document.formName.item("itemName") 这样的语句,不能在FF下运 ...

  9. jenkins配置QQ邮箱自动发送RF测试构建结果通知邮件

    声明:转载请注明出处,谢谢 首先确认QQ邮箱SMTP服务器的地址和端口号.如下图所示,请谨记,JENKINS全局邮箱配置需要使用: 步骤1:开启QQ邮箱的smtp服务:登陆QQ邮箱-设置-账户-开启P ...

  10. webpack.optimize.UglifyJsPlugin配置说明

    https://segmentfault.com/a/1190000008995453?utm_source=tuicool&utm_medium=referral