poj 3348 Cows

poj 3348 Cows

​​ 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

复习一下凸包算法 学了下如何去求面积 求凸包面积就是求出凸包之后把所有点按照凸包上的顺序两两求一下和起始点的三角形的面积 这个面积就是这两个向量叉积的二分之一

#include#include#includeusing namespace std;using namespace std;inline char gc(){ static char now[1<<16],*S,*T; if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;} return *S++;}inline int read(){ int x=0,f=1;char ch=gc(); while(!isdigit(ch)) {if(ch=='-') f=-1;ch=gc();} while(isdigit(ch)) x=x*10+ch-'0',ch=gc(); return x*f;}const int N=10010;struct node{ int x,y; inline friend node operator -(const node &a,const node &b){ return (node){a.x-b.x,a.y-b.y}; } inline friend int operator *(const node &a,const node &b){ return a.x*b.y-a.y*b.x; }}p[N],q[N];int n,top;inline int sqr(int x){return x*x;}inline int dis(const node &a,const node &b){ return sqr(a.x-b.x)+sqr(a.y-b.y);}inline bool cmp(const node &a,const node &b){ int tmp=(a-p[1])*(b-p[1]); if (!tmp) return dis(a,p[1])0;}int main(){ freopen("poj3348.in","r",stdin); n=read();int id=1; for (int i=1;i<=n;++i) p[i].x=read(),p[i].y=read(); for (int i=2;i<=n;++i) if (p[i].y>=1; printf("%d\n",ans/50); return 0;}

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