Educational Codeforces Round 64 (Rated for Div. 2)-C. Match Points-二分搜索

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题目传送门
题解：
Let’s denote the points that have greater coordinates in their matched pairs as R-points, and the points that have smaller coordinates as L-points.

Suppose we have an R-point that has smaller coordinate than some L-point. Then we can “swap” them, and the answer won’t become worse. Also, if some R-point has smaller coordinate than some point that doesn’t belong to any pair, or some L-point has greater coordinate than some point that doesn’t belong to any pair, we can swap them too. So, if the answer is k, we choose k leftmost points as L-points, and k rightmost ones as R-points.

For a fixed value of k, it’s easy to see that we should match the leftmost L-point with the leftmost R-point, the second L-point with the second R-point, and so on, in order to maximize the minimum distance in a pair. This fact allows us to check whether it is possible to construct at least k pairs, and we can use binary search to compute the answer to the problem.

利用算法导论证明贪心算法正确性的方法，复制粘贴，即可。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5;

int a[maxn+10];
int n = 0,z = 0;
int main(){
	while(~scanf("%d %d",&n,&z)){
		for(int i = 1;i <= n;++i){
			scanf("%d",&a[i]);
		}
		sort(a+1,a+1+n);
		int l = 0,r = n/2;
		int mid = 0,ans = 0;
		while(l <= r){
			mid = (l + r) / 2;
			int sum = 0;
			for(int i = 1;i <= mid;++i){
				if(a[n-mid+i] - a[i] >= z){
					++sum;
				}
			}
			if(sum == mid){
				ans = mid; 
				l = mid + 1;
			}else{
				r = mid - 1;
			}
		}
		printf("%d\n",ans);
	}
	
	return 0;
}