Cows
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 11278 | Accepted: 4948 |
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 101Sample Output
151解析：
计算几何：凸包+多边形面积。
关于求凸包大概就是先找到一个极坐标（瞎编的名字，就是一定在凸包上的点），以它为起点算出其他所有顶点的向量，按照一定的方向（顺时针）排序，然后贪心地加点与删点就行了。
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
using namespace std;
const int Max=1010;
int n,m,id=1,tot,ans;
struct point{
int x,y;
point(){}
point(const int &_x,const int &_y):x(_x),y(_y){}
friend point operator -(const point &a,const point &b){return point(a.x-b.x,a.y-b.y);}
friend int operator *(const point &a,const point &b){return a.x*b.y-a.y*b.x;}
inline int calc()const{return x*x-y*y;}
}p[Max],a[Max],q[Max];
inline int get_int()
{
int x=0,f=1;
char c;
for(c=getchar();(!isdigit(c))&&(c!='-');c=getchar());
if(c=='-') f=-1,c=getchar();
for(;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+c-'0';
return x*f;
}
inline bool comp(const point &a,const point &b)
{
int res=(a-p[1])*(b-p[1]);
if(res) return res>0;
return (a-p[1]).calc()<(b-p[1]).calc();
}
inline void init()
{
n=get_int();
for(int i=1;i<=n;i++)
{
p[i].x=get_int(),p[i].y=get_int();
if(p[id].x>p[i].x||(p[id].x==p[i].x&&p[id].y>p[i].y)) id=i;
}
if(id!=1) swap(p[id],p[1]);
sort(p+2,p+n+1,comp);
}
inline void solve()
{
q[++tot]=p[1];
for(int i=2;i<=n;i++)
{
while(tot>=3&&(p[i]-q[tot-1])*(q[tot]-q[tot-1])>0) tot--;
q[++tot]=p[i];
}
q[tot+1]=q[1];
for(int i=1;i<=tot;i++) ans+=q[i]*q[i+1];
printf("%d",ans/100);
}
int main()
{
init();
solve();
return 0;
}