poj 3348 Cow 凸包面积

Posted 2020-07-08 Przz

Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8122		Accepted: 3674

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

 

/*
poj 3348 Cow 凸包面积

给你n棵树绕成一个圈,问最多可以养多少牛
用所有点求出凸包然后求出多边形的面积即可

hhh-2016-05-08 18:47:21
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
using namespace std;
const int  maxn = 10100;
double PI = 3.1415926;
double eps = 1e-8;

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0)
        return -1;
    else
        return 1;
}

struct Point
{
    double x,y;
    Point() {}
    Point(double _x,double _y)
    {
        x = _x,y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x-b.x,y-b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y-y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};

struct Line
{
    Point s,t;
    Line() {}
    Line(Point _s,Point _t)
    {
        s = _s;
        t = _t;
    }
    pair<int,Point> operator &(const Line&b)const
    {
        Point res = s;
        if( sgn((s-t) ^ (b.s-b.t)) == 0)   //通过叉积判断
        {
            if( sgn((s-b.t) ^ (b.s-b.t)) == 0)
                return make_pair(0,res);
            else
                return make_pair(1,res);
        }
        double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
        res.x += (t.x-s.x)*ta;
        res.y += (t.y-s.y)*ta;
        return make_pair(2,res);
    }
};
Point lis[maxn];
int Stack[maxn],top;

double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
bool cmp(Point a,Point b)
{
    double t = (a-lis[0])^(b-lis[0]);
    if(sgn(t) == 0)
    {
        return dist(a,lis[0]) <= dist(b,lis[0]);
    }
    if(sgn(t) < 0)
        return false;
    else
        return true;
}

bool Cross(Point a,Point b,Point c)
{
   return  (b.y-a.y)*(c.x-b.x) == (c.y-b.y)*(b.x-a.x);
}

void Graham(int n)
{
    Point p;

    int k = 0;
    p = lis[0];
    for(int i = 1; i < n; i++)
    {
        if(p.y > lis[i].y || (p.y == lis[i].y && p.x > lis[i].x))
            p = lis[i],k = i;
    }
    swap(lis[0],lis[k]);
    sort(lis+1,lis+n,cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return ;
    }
    if(n == 2)
    {
        Stack[0] = 0,Stack[1] = 1;
        top = 2;
        return;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2; i < n; i++)
    {
        while(top > 1 && sgn((lis[Stack[top-1]]-lis[Stack[top-2]])
                             ^ (lis[i]-lis[Stack[top-2]])) <= 0)
            top --;
        Stack[top++] = i;
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n) != EOF)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf",&lis[i].x,&lis[i].y);
        }
        Graham(n);

        double res = 0;
        for(int i = 0;i < top;i++)
        {
            res += (lis[Stack[i]]^lis[Stack[(i+1)%top]])/2;
        }
        int ans = res;
        printf("%d\n",ans/50);
    }
    return 0;
}