POJ 2187：Beauty Contest 求给定一些点集里最远的两个点距离

Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 31414		Accepted: 9749

Description

Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farmers and their cows. For simplicity, the world will be represented as a two-dimensional plane, where each farm is located at a pair of integer coordinates (x,y), each having a value in the range -10,000 ... 10,000. No two farms share the same pair of coordinates. 

Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm 

Output

* Line 1: A single integer that is the squared distance between the pair of farms that are farthest apart from each other. 

Sample Input

4
0 0
0 1
1 1
1 0

Sample Output

2

Hint

Farm 1 (0, 0) and farm 3 (1, 1) have the longest distance (square root of 2) 

题意是给了二维坐标下的点，问在这些点中最远的两个点 距离的平方。

最远的两个点一定在凸包上，所以先找凸包上的点，在对凸包上的点求其距离平方找最大值。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int n;

struct no
{
	int x, y;
}node[50005];

int conbag[50005];
no orign;//原点

int dis(no n1, no n2)
{
	return (n1.x - n2.x)*(n1.x - n2.x) + (n1.y - n2.y)*(n1.y - n2.y);
}

int xmult(int x1, int y1, int x2, int y2)
{
	return x1*y2 - x2*y1;
}

int Across(no n1, no n2, no n3, no n4)
{
	return xmult(n2.x - n1.x, n2.y - n1.y, n4.x - n3.x, n4.y - n3.y);
}

int cmp(const void* n1, const void* n2)//qsort和我原来想象的是完全相反的。。。
{
	int temp = Across(orign, *(no *)n1, orign, *(no *)n2);

	if (temp > 0)
	{
		return -1;
	}
	else if (temp == 0)
	{
		return (dis(orign, *(no *)n1) - dis(orign, *(no *)n2));
	}
	else
	{
		return 1;
	}
}

int main()
{	
	int i, j, min_x, pos_x;
	while (cin >> n)
	{
		min_x = 10005;
		pos_x = 0;

		for (i = 1; i <= n; i++)
		{
			cin >> node[i].x >> node[i].y;

			if (node[i].x < min_x)
			{
				min_x = node[i].x;
				pos_x = i;
			}
			else if (min_x == node[i].x&&node[i].y < node[pos_x].y)
			{
				pos_x = i;
			}
		}
		orign = node[pos_x];
		qsort(node + 1, n, sizeof(no), cmp);

		int pc = 1;
		conbag[1] = 1;
		conbag[++pc] = 2;
		conbag[0] = 2;

		i = 3;
		while (i <= n)
		{
			if (Across(node[conbag[pc - 1]], node[conbag[pc]], node[conbag[pc]], node[i]) >= 0)
			{
				conbag[++pc] = i++;
				conbag[0]++;
			}
			else
			{
				pc--;
				conbag[0]--;
			}
		}
		int max_n = 0;
		for (i = 1; i <= conbag[0]; i++)
		{
			for (j = i + 1; j <= conbag[0]; j++)
			{
				max_n = max(dis(node[conbag[i]], node[conbag[j]]), max_n);
			}
		}
		cout << max_n << endl;
	}

	return 0;
}

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