Aggressive cows
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10078 | Accepted: 4988 |
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.
解析见代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
const int maxn=100000+100;
int a[maxn];
int n,c;
bool check(int dis)
{
int last=0;
for(int i=1;i<c;i++)//放置c头牛，共有c-1个空
{
int next=last+1;//next初始化
while(next<n&&a[next]-a[last]<dis) next++;//直到寻找到满足条件的next:
if(next>=n) return false;
last=next;//更新last
}
return true;
}
int main()
{
int ans;
while(scanf("%d%d",&n,&c)!=EOF)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);//使用二分逼近的方法确定答案，首先必须保证数组单调
int l=0,r=a[n-1]+1;//枚举的是距离，范围是0~a[n-1]+1;
while(l<=r)
{
int mid=(l+r)>>1;
if(check(mid)) ans=mid,l=mid+1;//满足条件，更新答案。
else r=mid-1;
}
printf("%d\n",ans);
}
return 0;
}
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
const int maxn=100000+100;
int a[maxn];
int n,c;
bool check(int dis)
{
int last=0;
for(int i=1;i<c;i++)//放置c头牛，共有c-1个空
{
int next=last+1;//next初始化
while(next<n&&a[next]-a[last]<dis) next++;//直到寻找到满足条件的next:
if(next>=n) return false;
last=next;//更新last
}
return true;
}
int main()
{
int ans;
while(scanf("%d%d",&n,&c)!=EOF)
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);//使用二分逼近的方法确定答案，首先必须保证数组单调
int l=0,r=a[n-1]+1;//枚举的是距离，范围是0~a[n-1]+1;
while(l<=r)
{
int mid=(l+r)>>1;
if(check(mid)) ans=mid,l=mid+1;//满足条件，更新答案。
else r=mid-1;
}
printf("%d\n",ans);
}
return 0;
}