poj 3348 Cows （凸包面积）

Cows

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 8861 Accepted: 3909

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

40 00 10175 075 101

Sample Output

151

Source

CCC 2007

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题目大意：给出一些树，可以选择任意的树圈地，每头牛需要50 square metres ，求最多放多少头牛。

题解：求凸包面积

叉积求面积

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define N 10003#define LL long longusing namespace std;struct vector {LL x,y;vector (LL X=0,LL Y=0){x=X,y=Y;}}p[N],ch[N];typedef vector point;vector operator -(vector a,vector b){return vector (a.x-b.x,a.y-b.y);}bool operator <(vector a,vector b){return a.x<b.x||a.x==b.x&&a.y<b.y;}int n,m;LL cross(vector a,vector b){return a.x*b.y-a.y*b.x;}void convexhull(point *p,int n,point *ch){sort(p+1,p+n+1);if (n==1) {ch[m++]=p[1];return ;}m=1;for (int i=1;i<=n;i++){while (m>2&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<0) m--;ch[m++]=p[i]; }int k=m;for (int i=n-1;i>=1;i--){while (m>k&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<0) m--;ch[m++]=p[i];}m--;}LL Abs(LL x){if (x<0) return -x;else return x;}int main(){freopen("a.in","r",stdin);while (scanf("%d",&n)!=EOF){for (int i=1;i<=n;i++) scanf("%lld%lld",&p[i].x,&p[i].y);if (n<3) {printf("0\n");continue;}convexhull(p,n,ch);LL area=0; m--;//for (int i=1;i<=m;i++) cout<<ch[i].x<<" "<<ch[i].y<<endl;for (int i=2;i<=m-1;i++) area+=Abs(cross(ch[i]-ch[1],ch[i+1]-ch[1]));printf("%lld\n",(LL)area/100);    }}