POJ - 3685 Matrix

POJ - 3685 Matrix

二分套二分

Matrix - POJ 3685 - Virtual Judge (vjudge.net)

\(f(i,j)=i^2+10^5*i+j^2-10^5*j+i*j\)

\(f\) 关于 \(i\) 单调递增，关于 \(j\) 单调递减

要找第 \(m\) 小的 $ f(i,j)$, 则可二分答案，check 函数返回矩阵中有多少元素小于当前的二分值

check 中求有多少元素小于某个值的过程中，可以枚举 \(j\), \(i\) 单调递增，所以可以二分出每个 \(j\) 对应有多少个 \(i\) 的 \(f(i,j)\) 小于当前的二分值，

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

using namespace std;
typedef long long ll;
const ll N = 1e5;
int n;
ll m;

ll f(ll x, ll y)
{
	return x * x + N * x + y * y - N * y + x * y;
}

ll check(ll x)
{
	ll cnt = 0;
	for (int j = 1; j <= n; j++)
	{
		int l = 0, r = n + 1;
		while(l + 1 != r)
		{
			int mid = l + r >> 1;
			if (f(mid, j) < x)
				l = mid;
			else
				r = mid;
		}
		cnt += l;
	}
	return cnt;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int T;
	cin >> T;
	while(T--)
	{
		cin >> n >> m;
		ll l = -1e12, r = 1e12;
		while(l + 1 != r)
		{
			ll mid = l + r >> 1;
			if (check(mid) < m)
				l = mid;
			else
				r = mid;
		}
		cout << l << endl;
	}
	return 0;
}