POJ 2456 Aggressive cows (二分 基础)
来源:互联网 发布:遗传算法基本思想 编辑:程序博客网 时间:2024/05/22 12:15
Aggressive cows
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7924 Accepted: 3959
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?
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
* 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 312849
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.
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.
Source
USACO 2005 February Gold
题目链接:http://poj.org/problem?id=2456
题目大意:一个数轴上n个点,每个点一个整数值,有c个物品,要放在这些点的某几个上,求怎么放可以使任意两个物品间距离的最小值最大,求这个最大值
题目分析:最小值最大,典型二分题,二分距离的值判断
题目链接:http://poj.org/problem?id=2456
题目大意:一个数轴上n个点,每个点一个整数值,有c个物品,要放在这些点的某几个上,求怎么放可以使任意两个物品间距离的最小值最大,求这个最大值
题目分析:最小值最大,典型二分题,二分距离的值判断
#include <cstdio>#include <algorithm>using namespace std;int const INF = 0x3fffffff;int const MAX = 1e5 + 5;int d[MAX];int n, c;int cal(int m){int ans = 0, now = d[0];for(int i = 1; i < n; ){while(d[i] < now + m)i ++;ans ++;now = d[i];}return ans;}int main(){scanf("%d %d", &n, &c);for(int i = 0; i < n; i++)scanf("%d", &d[i]);sort(d, d + n);int r = d[n - 1] - d[0], l = 0, ans = 0;while(l <= r){int m = (l + r) / 2;int num = cal(m);if(num >= c){ans = m;l = m + 1;}elser = m - 1;}printf("%d\n", ans);}
0 0
- POJ 2456 Aggressive cows (二分 基础)
- POJ 2456 Aggressive cows 二分
- POJ 2456 Aggressive cows 二分
- poj 2456 Aggressive cows (二分)
- poj 2456 Aggressive cows 二分
- POJ - 2456 Aggressive cows 二分
- POJ 2456 Aggressive cows 【二分】
- POJ 2456 Aggressive cows(二分)
- POJ 2456 Aggressive cows 二分
- poj 2456 二分 Aggressive cows
- [POJ 2456]Aggressive cows[二分]
- 【POJ 2456 Aggressive cows】+ 二分
- poj 2456 Aggressive cows (二分)
- POJ 2456 Aggressive cows 二分
- poj 2456 Aggressive cows poj 3258 (二分)
- POJ 2456 Aggressive cows(二分)
- POJ 2456 - Aggressive cows(二分)
- poj 2456 Aggressive cows,二分,最大化最小值
- 网络IO模型:同步IO和异步IO,阻塞IO和非阻塞IO
- hdu 4308 Saving Princess claire(BFS)
- 【实训项目】--《手机通讯录》
- 数组
- [CortexM0--stm32f0308]clock介绍
- POJ 2456 Aggressive cows (二分 基础)
- python serial安装模块
- [Python]django使用多进程连接msyql错误
- 随手可得的Application对象
- struts (四) path DMI
- YUM LINUX安装ftp
- 【分块】 CF 551 E GukiZ and GukiZiana
- GDB十分钟教程
- Jquery 获得输入框文件名和后缀名