POJ 2456 Aggressive cows (二分 基础)
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Aggressive cows
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7924 | Accepted: 3959 |
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.
Source
USACO 2005 February Gold
题目链接:http://poj.org/problem?
题目链接:http://poj.org/problem?
id=2456
题目大意:一个数轴上n个点,每一个点一个整数值,有c个物品。要放在这些点的某几个上,求怎么放能够使随意两个物品间距离的最小值最大,求这个最大值
题目分析:最小值最大,典型二分题。二分距离的值推断
#include <cstdio> #include <algorithm> using namespace std; int const INF = 0x3fffffff; int const MAX = 1e5 + 5; int d[MAX]; int n, c; int cal(int m) { int ans = 0, now = d[0]; for(int i = 1; i < n; ) { while(d[i] < now + m) i ++; ans ++; now = d[i]; } return ans; } int main() { scanf("%d %d", &n, &c); for(int i = 0; i < n; i++) scanf("%d", &d[i]); sort(d, d + n); int r = d[n - 1] - d[0], l = 0, ans = 0; while(l <= r) { int m = (l + r) / 2; int num = cal(m); if(num >= c) { ans = m; l = m + 1; } else r = m - 1; } printf("%d\n", ans); }
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