POJ 3348：Cows 凸包+多边形面积

Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7739		Accepted: 3507

Description

Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.

However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.

Input

The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and y separated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).

Output

You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.

Sample Input

4
0 0
0 101
75 0
75 101

Sample Output

151

题意是给出各个树的位置，用这些树可以围出一块最大的面积，一头牛需要50平方米的面积，问这些树围出的面积可以养多少头牛。

先构造凸包，在求出凸包围成的面积，最后除以50求整。

代码：

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

struct no
{
	int x, y;
}node[10005];

int n;
int conbag[10005];

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)
{
	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, k, min_x, pos_x;
	double sum;
	while (scanf("%d", &n) != EOF)
	{
		min_x = 1005;
		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]--;
			}
		}

		if (n < 3 || conbag[0]<3)
		{
			cout << 0 << endl;
		}
		else
		{
			sum = 0;
			for (i = 1; i + 1 <= conbag[0]; ++i)
			{
				sum += abs((node[conbag[i]].x * node[conbag[(i + 1)]].y - node[conbag[i]].y * node[conbag[(i + 1)]].x));
			}
			sum += abs((node[conbag[i]].x * node[1].y - node[conbag[i]].y * node[1].x));
			sum = sum / 2.0;
			cout << (int)(sum / 50) << endl;
		}
	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。