POJ 25456 Aggressive cows 【二分最大化最小值】
阿新 • • 發佈：2019-02-11
Aggressive cowsTime Limit:1000MS Memory Limit:65536KB 64bit
IO Format:%lld & %llu
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?
* Lines 2..N+1: Line i+1 contains an integer stall location, xi
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.
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 distanceSample 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.
題目描述：
農夫約翰搭了一間牛棚，有N個牛舍。牛舍排在一條線上，第i號牛舍在xi的位置。但是他的M頭牛對小屋不滿意，因此經常相互攻擊。約翰為了防止牛之間相互傷害，
因此決定把每頭牛都放在離其他牛儘可能遠的牛舍。也就是要最大化最近的兩頭牛之間的距離。
<span style="font-size:18px;">#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int a[100010]; int main() { int n,m; while(~scanf("%d %d",&n, &m)) { for(int i=0; i<n; i++) { scanf("%d",&a[i]); } sort(a,a+n); int l=0,r=a[n-1]; int ans=0,sum=0; while(l<=r){ int mid=(l+r)>>1; int ant=a[0]; ans=1; for(int i=1;i<n;i++){ if(a[i]-ant>=mid){ ans++; ant=a[i]; } } if(ans>=m){ sum=max(sum,mid); l=mid+1; } else r=mid-1; } printf("%d\n",sum); } return 0; } </span>