CodeForces - 1730B Meeting on the Line

CodeForces - 1730B Meeting on the Line

n people live on the coordinate line, the �i-th one lives at the point ��xi​ (1≤�≤�1≤i≤n). They want to choose a position �0x0​ to meet. The �i-th person will spend ∣��−�0∣∣xi​−x0​∣ minutes to get to the meeting place. Also, the �i-th person needs ��ti​ minutes to get dressed, so in total he or she needs ��+∣��−�0∣ti​+∣xi​−x0​∣ minutes.

Here ∣�∣∣y∣ denotes the absolute value of �y.

These people ask you to find a position �0x0​ that minimizes the time in which all �n people can gather at the meeting place.

Input

The first line contains a single integer �t (1≤�≤1031≤t≤103) — the number of test cases. Then the test cases follow.

Each test case consists of three lines.

The first line contains a single integer �n (1≤�≤1051≤n≤105) — the number of people.

The second line contains �n integers �1,�2,…,��x1​,x2​,…,xn​ (0≤��≤1080≤xi​≤108) — the positions of the people.

The third line contains �n integers �1,�2,…,��t1​,t2​,…,tn​ (0≤��≤1080≤ti​≤108), where ��ti​ is the time �i-th person needs to get dressed.

It is guaranteed that the sum of �n over all test cases does not exceed 2⋅1052⋅105.

Output

For each test case, print a single real number — the optimum position �0x0​. It can be shown that the optimal position �0x0​ is unique.

Your answer will be considered correct if its absolute or relative error does not exceed 10−610−6. Formally, let your answer be �a, the jury's answer be �b. Your answer will be considered correct if ∣�−�∣���(1,∣�∣)≤10−6max(1,∣b∣)∣a−b∣​≤10−6.

两种思路解题

（1）

  贪心思想：

#include<bits/stdc++.h>
#define Acode ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
#define endl '\n'
const int inf = 0x3f3f3f3f;

const int N = 1e6 + 10;
ll x[N], t1[N];

int main()
{
	Acode;
	int t;
	cin >> t;
	while (t--)
	{
		int n;
		cin >> n;
		for (int i = 1; i <= n; i++) cin >> x[i];
		for (int i = 1; i <= n; i++)cin >> t1[i];

		ll min1 = inf, max1 = -inf;
		for (int i = 1; i <= n; i++)
		{
			min1 = min(min1, x[i] - t1[i]);
			max1 = max(max1, x[i] + t1[i]);
		}
		printf("%.10f\n", (min1 + max1) / 2.0);

	}
	return 0;
}

（2）

 二分思想

#include<bits/stdc++.h>
#define Acode ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
#define endl '\n'
double ans = 0;
const int N = 1e6 + 10;
ll a[N], t[N];
int n;
bool check(double mid)
{
	double l = -inf, r = inf;
	for (int i = 1; i <= n; i++)
	{
		l = max(l, a[i] + t[i]-mid);
		r = min(r, a[i] + mid - t[i]);
	}
	if (l <= r)
	{
		ans = (l + r) / 2;
		return true;
	}
	else return false;
}

int main()
{
	Acode;
	int t1;
	cin >> t1;
	while (t1--)
	{
		cin >> n;
		for (int i = 1; i <= n; i++) cin >> a[i];
		for (int i = 1; i <= n; i++) cin >> t[i];
		double l = 0, r = 1e9;
		int cnt = 100;
		while (cnt--)
		{
			double mid = (l + r) / 2;
			if (check(mid)) l = mid;
			else r = mid;
		}
		printf("%.10f\n", ans);
	}

	return 0;
}