POJ 2456 Aggressive cows (二分 最大化最小值)

 /*
 qq:1239198605
 ctgu_yyf
        */

#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;

int x[100005];
int n,m;

bool c(int d)
{
	int last = 0;
	for(int i=1; i<m; i++)
	{
		int crt=last+1;
		while(crt<n&&x[crt]-x[last]<d)
		{
			crt++;
		}
		if(crt==n)
		return false;
		
		last=crt;
	}
	return true;
}

void solve()
{
	//对x数组进行排序 
	sort(x,x+n);
	
	//初始化解的存在范围
	int lb=0,rb=x[n-1]-x[0];
	
	while(rb-lb>1)
	{
		int mid=(lb+rb)/2;
		if(c(mid)) lb=mid;
		else
		rb=mid;
		
		
	}	 
	cout<<lb<<endl;
}

int main()
{
   ios::sync_with_stdio(false);
   
   cin>>n>>m;
   for(int i=0;i<n;i++)
   cin>>x[i];
   
   solve();


return 0;
}

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转载自blog.csdn.net/k_koris/article/details/81665933