二分贪心05
来源:互联网 发布:初创公司如何管理 知乎 编辑:程序博客网 时间:2024/06/05 03:10
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.
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.
这是典型的例题,求最大的最小距离,直接代码
#include <iostream>#include<stdio.h>#include<algorithm>using namespace std;int n,c;int a[100010];int judge(int x)//判断当前的mid是否可以住的下这些牛{ int cnt=1; int tem=a[1]; for(int i=2;i<=n;++i) { if(a[i]-tem>=x) { cnt++; tem=a[i]; } } if(cnt>=c) return 0; else return 1;}int main(){ while(cin>>n>>c) { for(int i=1;i<=n;++i) scanf("%d",&a[i]); sort(a,a+1+n); int left=0,mid,right=a[n]-a[1]; while(left<=right) { mid=(left+right)/2; if(judge(mid))//住不下 right=mid-1; else//住下 left=mid+1; } printf("%d\n",left-1); } return 0;}
0 0
- 二分贪心--05
- 二分贪心05
- ACM-二分贪心E-05
- 二分+贪心
- 贪心 + 二分
- 贪心二分
- 贪心+二分
- 贪心(bnuoj49103+二分+贪心)
- BNU 49103 贪心【二分+贪心】
- hdu pie(二分+贪心)
- POJ3497 Assemble 二分+贪心
- mysterious 二分加贪心
- HDU 3650 贪心+二分
- 11627 - Slalom (二分+贪心)
- 二分+贪心+LA3177
- hdu4004(二分+贪心)
- hdu1677(贪心+二分)
- poj2456(贪心+二分)
- Eclipse加载插件失败
- Leetcode:Word Ladder
- 运维_win server2008关闭危险端口445,135,137,138,139的方法
- Tensorflow实现卷积神经网络模型
- 解决jad.exe查看外部jar包源码失败问题
- 二分贪心05
- Quartz 2D编程指南(6) - 阴影(Shadows)
- C/C++:switch语句
- java基础之—类加载器
- 古北水镇上空的夜景,怎少的了S8的“夜视仪”?
- 实验四 DPCM编码
- 动态规划算法——知识点总结
- Quartz 2D编程指南(7) - PDF文档的创建、显示及转换
- 从斜率DP讲起