Aggressive cows
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4808 | Accepted: 2356 |
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 3 1 2 8 4 9
Sample Output
3
求解距离最小值可以取到的最大值（有点乱呃……就是最小值的最大化问题）
求解的方法依然是二分搜索，和POJ 1064差不多
首先读入所有的x[i]（0<=i<=n-1）后进行从小到大的排序，即可确定解区间为1~x[n-1]
然后在解区间内用二分搜索的方法找到最终的答案。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 6 using namespace std; 7 8 int n,c; 9 int x[100010]; 10 11 int main() 12 { 13 scanf("%d %d",&n,&c); 14 for(int i=0;i<n;i++) 15 scanf("%d",&x[i]); 16 sort(x,x+n); 17 int l=1,r=x[n-1]; 18 while(r-l>1) 19 { 20 int mid=(l+r)/2,ans=0; 21 for(int i=0;ans<c&&i<n;) 22 { 23 int t=i; 24 while(x[++i]-x[t]<mid); 25 ans++; 26 } 27 if(ans>=c) 28 l=mid; 29 else 30 r=mid; 31 } 32 if(l!=r) 33 { 34 int ans=0; 35 for(int i=0;ans<c&&i<n;) 36 { 37 int t=i; 38 while(x[++i]-x[t]<r); 39 ans++; 40 } 41 if(ans>=c) 42 l=r; 43 } 44 printf("%d ",l); 45 46 return 0; 47 }