POJ：2456-Aggressive cows

Aggressive cows

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 18313 Accepted: 8716

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 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.

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解题心得：

1. 有n间小屋，n个小屋排在一列，每个小屋距离原点有一个距离，要将m头牛放在小屋中，要使两头牛之间的距离尽量的大，如果最大化牛直接的距离，那么最近两头牛之间的距离最大是多少。
2. 看到这个最大化最小值那么首先想到的就是二分，最大之间求最小，最小之间求最大，平均之间求最大都是二分。每次二分最大距离，然后按照这个距离安排，看能否用n个小屋安排下，不能那么减少距离，如果可以那么增大距离。

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#include <algorithm>
#include <cstring>
#include <stdio.h>
using namespace std;
const int maxn = 1e5+10;
int pos[maxn],n,c;

void init() {
    for(int i=0;i<n;i++)
        scanf("%d",&pos[i]);
    sort(pos,pos+n);
}

bool checke(int len){
    int cnt = 1,now = pos[0];
    for(int i=1;i<n;i++) {
        if (pos[i] - now >= len) {
            cnt++;
            now = pos[i];
        }
    }
    return cnt >= c;
}

int binary_search() {
    int l = 0,r = 1e9+10;
    while(r-l > 1) {
        int mid = (l+r)/2;
        if(checke(mid))
            l = mid;
        else
            r = mid;
    }
    return l;
}

int main() {
    while(scanf("%d%d",&n,&c) != EOF) {
        init();
        int ans = binary_search();
        printf("%d\n",ans);
    }
    return 0;
}