ZOJ 3194 Coverage【贪心】【思维题】

Coverage

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

2
3
1 1
2 2
3 3

3
1 2
4 1
2 5

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
", ans / 2.0);
	}
	return 0;
}