POJ - 3579 Median

POJ - 3579 Median

二分套二分

Median - POJ 3579 - Virtual Judge (vjudge.net)

对数轴上的点排序后，可枚举左端点，固定左端点后，区间长度关于右端点是单调递增的，因此可以二分

1. 二分答案
2. check 返回有多少区间小于当前的二分值，求有多少区间的过程可以枚举左端点，二分右端点

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int a[N];
int n;
ll m;

ll check(int x)
{
	ll cnt = 0;
	for (int i = 1; i < n; i++)
	{
		int l = i, r = n + 1;
		while(l + 1 != r)
		{
			int mid = l + r >> 1;
			if (a[mid] - a[i] < x)
				l = mid;
			else
				r = mid;
		}
		cnt += l - i;
	}
	return cnt;
}
int main()
{
	while(scanf("%d", &n) != EOF)
	{
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		sort(a + 1, a + n + 1);
		m = 1ll * n * (n - 1) / 2;
		m = (m + 1) / 2;
		int l = 0, r = 1e9 + 1;
		while(l + 1 != r)
		{
			int mid = l + r >> 1;
			if (check(mid) < m)//比mid小的有几个
				l = mid;
			else
				r = mid;
		}
		printf("%d\n", l);
	}
	return 0;
}