ZOJ 3194 Coverage【贪心】【思维题】

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本文介绍了一个算法问题，即如何通过调整一组给定点中 y 坐标的顺序来最大化由这些点构成的线段下方的覆盖面积。文章提供了一段 C++ 代码实现，通过排序 x 和 y 坐标并计算每一对相邻 x 值之间的距离来解决该问题。

Coverage

Time Limit: 1 Second Memory Limit: 32768 KB

Given n points (x_i, y_i) (i = 1, 2 ... n) in a plane, where all x_i will be distinct. After connecting the n points with staight lines in order from the leftmost point to the rightmost point, the area below the lines is called coverage. Now, it's your job to calculate the maximum coverage area, if y_i (i = 1, 2 ... n) can be swapped arbitrarily.

Input

The first line of the input is an integer t (t <= 100), indicate the number of cases.

Each case starts with one integer n (2 <= n <= 1000) in a line. Then follows n lines, each consists of two integers x y (1 <= x, y <= 10⁵) representing a point.

Cases are separated by one blank line.

Output

For each case, output the answer in a line, keep 1 digit after the decimal point.

Sample Input

Sample Output

4.5
10.0

题意：对于题目给的点，x固定，而与x组合的y可以任意交换，求如何安置y可使这些点组成线段下面的面积最大，最大面积是多少

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;

const int mod = 1e9 + 7;
const int maxn = 1010;

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n;
		double x[maxn], y[maxn], dis[maxn];
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
		{
			scanf("%lf%lf", x + i, y + i);
		}
		sort(x, x + n);
		sort(y, y + n);
		dis[0] = x[1] - x[0];
		dis[n - 1] = x[n - 1] - x[n - 2];
		for (int i = 1; i < n - 1; i++)
		{
			dis[i] = x[i + 1] - x[i - 1];
		}
		sort(dis, dis + n);
		double ans = 0;
		for (int i = 0; i < n; i++)
		{
			ans += dis[i] * y[i];
		}
		printf("%.1f\n", ans / 2.0);
	}
	return 0;
}