poj3348 Cows

Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10039		Accepted: 4408

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

Source

CCC 2007

大致题意：n个点，用一个绳子圈出尽可能大的面积，养一头牛要50的面，最多能养多少头？

分析：凸包模板题.几个要点：1.找极点，将序号设为1. 2.根据极角和极点的距离排序. 3.维护一个最少有2个点的栈，再来考虑每个点能不能加进来，加进来会将多少个点从栈中删除，利用叉积判断.  4.利用叉积求面积.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

int n,tot,ans;

struct node
{
    int x,y;
}p[10010],q[10010];

node sub(node a,node b)
{
    node temp;
    temp.x = a.x - b.x;
    temp.y = a.y - b.y;
    return temp;
}

int det(node a,node b)
{
    return a.x * b.y - a.y * b.x;
}

int dist(node a)
{
    return (a.x - p[1].x) * (a.x - p[1].x) + (a.y - p[1].y) * (a.y - p[1].y);
}

bool cmp(node a,node b)
{
    int temp = det(sub(a,p[1]),sub(b,p[1]));
    if (temp != 0)
        return temp > 0;
    return dist(a) < dist(b);
}

void Graham()
{
    int id = 1;
    for (int i = 2; i <= n; i++)
    {
        if (p[i].x < p[id].x || (p[i].x == p[id].x && p[i].y < p[id].y))
            id = i;
    }
    if (id != 1)
        swap(p[1],p[id]);
    sort(p + 2,p + 1 + n,cmp);
    q[++tot] = p[1];
    for (int i = 2; i <= n; i++)
    {
        while (tot >= 2 && det(sub(p[i],q[tot - 1]),sub(q[tot],q[tot - 1])) >= 0)
            tot--;
        q[++tot] = p[i];
    }
}

int solve()
{
    int res = 0;
    q[tot + 1] = q[1];
    for (int i = 1; i <= tot; i++)
        res += det(q[i],q[i + 1]);
    return res / 2;
}

int main()
{
    scanf("%d",&n);
    for (int i = 1; i <= n; i++)
        scanf("%d%d",&p[i].x,&p[i].y);
    Graham();
    ans = solve();
    printf("%d\n",ans / 50);

    return 0;
}