疯牛-- Aggressive cows (二分)
疯牛
- 描述
- 农夫 John 建造了一座很长的畜栏,它包括N (2 <= N <= 100,000)个隔间,这些小隔间依次编号为x1,...,xN (0 <= xi <= 1,000,000,000).
但是,John的C (2 <= C <= N)头牛们并不喜欢这种布局,而且几头牛放在一个隔间里,他们就要发生争斗。为了不让牛互相伤害。John决定自己给牛分配隔间,使任意两头牛之间的最小距离尽可能的大,那么,这个最大的最小距离是什么呢?
- 输入
- 有多组测试数据,以EOF结束。 第一行:空格分隔的两个整数N和C 第二行——第N+1行:分别指出了xi的位置
- 输出
- 每组测试数据输出一个整数,满足题意的最大的最小值,注意换行。
- 样例输入
-
5 3
1
2
8
4
9 - 样例输出
-
3
题解:二分;
代码:#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXN=1e5+;
int m[MAXN];
int N,C;
bool js(int x){
int t=m[];
int cnt=;
for(int i=;i<N;i++){
if(m[i]-t>=x)cnt++,t=m[i];
}
if(cnt>=C)return true;
else return false;
}
int erfen(int l,int r){
int mid;
while(l<=r){
mid=(l+r)/;
if(js(mid))l=mid+;
else r=mid-;
}
return l;
}
int main(){
while(~scanf("%d%d",&N,&C)){
for(int i=;i<N;i++)scanf("%d",m+i);
sort(m,m+N);
printf("%d\n",erfen(,m[N-]-m[])-);//不明觉厉
}
return ;
}Aggressive cowsTime Limit: 1000MS Memory Limit: 65536K Total Submissions: 9080 Accepted: 4512 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, xiOutput
* Line 1: One integer: the largest minimum distanceSample Input
5 3
1
2
8
4
9Sample 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.
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