#题解#洛谷P3143

[传送门](P3143 [USACO16OPEN] Diamond Collector S - 洛谷)

分析

解决问题只需枚举这样的两个组：使得每组中极差<=k，且两个组不交，求两个组元素

代码实现

#include<bits/stdc++.h>
using namespace std;
int k, n;
const int N = 1e5+10;
int a[N], q[N];
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> k;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	sort(a + 1, a + 1 + n);
	
	for (int i = 1; i <= n; i++)
	{
		int j = i;
		while (a[i] + k >= a[j] && j <= n)
			j++;
		q[i] = j;//[i,j-1]是为一组
	}
	int ans = -1;
	//枚举两个组
	for (int i = 1; i <= n; i++)
		for (int j = q[i]; j <= n; j++)
			ans = max(ans, q[i] - i + q[j] - j);
	cout << ans;

	return 0;
}

Trick/错误 总结

1.双指针查询一维数组的连续子段

2.ans中两组不交