Aggressive cows
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 23866 | Accepted: 11141 |
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
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.
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.
尝试了剪枝 结果运行时间并没有什么区别(注释掉的地方是剪枝)
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
const int si = 100010, inf = 1000000000 + 10;
using namespace std;
int N, COW;
int ar[si];
bool c(int dis) {
int pre = 0, finished = 1;
for (int i = 1; i < N && finished < COW; i++) {
//if (ar[N - 1] - ar[pre] < dis * (COW - finished)) return false;
//剪枝 剩下COW - finished头牛 每头牛需要dis的距离
if (ar[i] - ar[pre] >= dis) {
finished++;
pre = i;
}
}
return finished >= COW;
}
int main() {
cin >> N >> COW;
for (int i = 0; i < N; i++) scanf("%d", &ar[i]);;
sort(ar, ar + N);
int l = 1, m, r = inf;
while (l < r - 1) {
m = l + r >> 1;
if (c(m)) l = m;
else r = m;
}
cout << l << endl;
return 0;
}
