Poj 2456 Aggressive cows ( 二分+贪心

Aggressive cows

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 distance

Sample Input

5 312849

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.

题意

给N间房子，最大化最小值 求每个奶牛的最大距离

题解：

二分求最大化最小值
先贪心 将符合情况的d判断出来

先对房子排序

把第一头牛放进第一个房间内

将第i头牛扔进房子 符合Xi+1 - Xi > d 寻找最大的Xi+1

然后二分查找所有情况

AC代码

#include <cstdio>#include <algorithm>using namespace std;#define N 100010int arr[N];int n, m;bool check(int d)   // check d的值 是否合理{    int f = 0;    for(int i = 1;i < m; i++) {        int cur = f+1;        while(cur<n && arr[cur]-arr[f]<d) {            cur++;        }        if(cur == n) return false;        f = cur;    }return true; } int main(){    scanf("%d%d",&n,&m);    for(int i = 0;i < n; i++)         scanf("%d",&arr[i]);    sort(arr,arr+n);    int low = 0; int high = arr[n-1]-arr[0];    while(high>=low) {        int mid = (low+high) >> 1;        if(check(mid)) low = mid+1;        else high = mid-1;    }    printf("%d\n",low-1);return 0;}