Cows

Time Limit: 2000MS Memory Limit: 65536K

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

很显然，这道题是要让我们算出给出点的凸包的面积。综上所述：这是一道凸包的裸板题，我本蒟蒻用的是Graham" role="presentation" style="position: relative;">GrahamGraham扫描法。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 10005
using namespace std;
struct pot{
    double x,y;
}p[N],vec[N];
inline double cross(pot a,pot b){return a.x*b.y-a.y*b.x;}
inline pot Vector(pot a,pot b){pot c;c.x=a.x-b.x,c.y=a.y-b.y;return c;}
inline double calc(pot *p,int n){
    double ans=0;
    for(int i=1;i<n-1;++i)
        ans+=cross(Vector(p[i],p[0]),Vector(p[i+1],p[0]));
    return ans/2;
}
inline bool cmp(const pot&x,const pot&y){return x.x==y.x?x.y<=y.y:x.x<y.x;}
inline int solve(pot *p,int n,pot *vec){
    sort(p,p+n,cmp);
    int m=0;
    for(int i=0;i<n;++i){
        while(m>1&&cross(Vector(vec[m-1],vec[m-2]),Vector(p[i],vec[m-2]))<=0)--m;
        vec[m++]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;--i){
        while(m>k&&cross(Vector(vec[m-1],vec[m-2]),Vector(p[i],vec[m-2]))<=0)--m;
        vec[m++]=p[i];
    }
    if(n>1)m--;
    return m;
}
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;++i)scanf("%lf%lf",&p[i].x,&p[i].y);
    int m=solve(p,n,vec),ans=0;
    double area=calc(vec,m);
    while(area>=50)area-=50,++ans;
    printf("%d",ans);
}