POJ2456 Aggressive cows
阿新 • • 發佈:2018-12-30
Aggressive cows
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?
* Line 1: Two space-separated integers: N and C
* 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.
Time Limit: 1000MS | Memory Limit: 65536K |
Total Submissions: 6510 | Accepted: 3263 |
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
* 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.
Source
二分列舉
#include <stdio.h> #include <string.h> #include <algorithm> #define maxn 100010 int pos[maxn], N, C; bool cal(int k) { int i, m, cnt = 0, s, t; s = 0; for(m = 1; m < C; ++m) { t = s + 1; while(t < N && pos[t] - pos[s] < k) ++t; if(t == N) return false; s = t; } return true; } int main() { int i, left, mid, right; while(scanf("%d%d", &N, &C) != EOF) { for(i = 0; i < N; ++i) scanf("%d", &pos[i]); std::sort(pos, pos + N); left = 0; right = pos[N - 1] + 1; while(right - left > 1) { mid = (left + right) >> 1; if(cal(mid)) left = mid; else right = mid; } printf("%d\n", left); } return 0; }