[二分+贪心] poj 2456 Aggressive cows_agressive cows 数算python-优快云博客

本文链接：https://blog.youkuaiyun.com/tlonline/article/details/47187089

本文解决了一个经典的算法问题——如何在给定的牛棚位置中放置一定数量的牛，使得任意两只牛之间的最小距离尽可能大。通过使用二分查找和贪心策略来找到最优解。

N - Aggressive cows
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit

Status
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.

还是二分，上界INF,下界为0。
check(d) 表示是否可以安排牛的位置使得最近的两头牛的距离不小于d。
很容易用贪心求解：
1、对牛舍的位置进行排序
2、把第一头牛放入x0的牛舍
3、将下一头牛放入满足距离刚好大于或等于d的上一头牛的牛舍中

#include <iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
const int MAXN = 100010;
const int INF =0X3f3f3f3f;
int N,M;
int x[MAXN];
bool check(int d)
{
    int last = 0;
    for(int i = 1; i < M; i++)
    {
        int now = last + 1;
        while(now < N && x[now]- x[last] < d)
            now++;
        if (now == N) return false;
        last = now;
    }
    return true;

}

int main()
{
    scanf("%d%d",&N,&M);
    for(int i = 0; i < N;i++)
        scanf("%d",&x[i]);
    sort(x,x+N);
    int lb = 0,ub = INF;
    while(ub - lb > 1)
    {
        int mid = (lb + ub) /2;
        if (check(mid)) lb = mid;
        else ub = mid;
    }
    printf("%d\n",lb);
    return 0;
}