poj 3348:Cows（计算几何，求凸包面积）

Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6199		Accepted: 2822

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

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　　计算几何，求凸包面积。

　　题意是给你n个点，让你求包围这n个点的凸包面积。求凸包面积的思路是先求凸包，再求凸包构成的多边形面积。

　　根据题意最后别忘了除以50。为了求速度，可耻的直接套用了模板，改了改就可耻过了。

　　代码：

 1 #include <iostream>

 2 #include <iomanip>

 3 #include <cmath>

 4 #include <stdio.h>

 5 using namespace std;  6 

 7 struct Point{  8     double x,y;  9 }; 10 double dis(Point p1,Point p2) 11 { 12     return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); 13 } 14 double xmulti(Point p1,Point p2,Point p0)    //求p1p0和p2p0的叉积,如果大于0,则p1在p2的顺时针方向

15 { 16     return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 17 } 18 int graham(Point p[],int n,int pl[])    //求凸包，返回凸包中点的个数+1

19 { 20     //找到纵坐标（y）最小的那个点，作第一个点 

21     int i; 22     int t = 1; 23     for(i=1;i<=n;i++) 24         if(p[i].y < p[t].y) 25             t = i; 26     pl[1] = t; 27     //顺时针找到凸包点的顺序，记录在 int pl[]

28     int num = 1;    //凸包点的数量

29     do{    //已确定凸包上num个点 

30         num++; //该确定第 num+1 个点了

31         t = pl[num-1]+1; 32         if(t>n) t = 1; 33         for(int i=1;i<=n;i++){    //核心代码。根据叉积确定凸包下一个点。 

34             double x = xmulti(p[i],p[t],p[pl[num-1]]); 35             if(x<0) t = i; 36  } 37         pl[num] = t; 38     } while(pl[num]!=pl[1]); 39     return num-1; 40 } 41 double getS(Point a,Point b,Point c)    //返回三角形面积 

42 { 43     return ((b.x - a.x) * (c.y - a.y) - (b.y - a.y)*(c.x - a.x))/2; 44 } 45 double getPS(Point p[],int pl[],int n)    //返回多边形面积。必须确保 n>=3，且多边形是凸多边形 

46 { 47     double sumS=0; 48     for(int i=2;i<=n-1;i++) 49         sumS+=fabs(getS(p[pl[1]],p[pl[i]],p[pl[i+1]])); 50     return sumS; 51 } 52 int main() 53 { 54     int n; 55     while(cin>>n){ 56         Point p[10005]; 57         int pl[10005]; 58         int i; 59         for(i=1;i<=n;i++)    //输入点

60             scanf("%lf%lf",&p[i].x,&p[i].y); 61         int num = graham(p,n,pl);    //求凸包，并返回凸包点个数

62         printf("%d\n",int(getPS(p,pl,num)/50));    //求凸包面积

63  } 64     return 0; 65 }

 

Freecode : www.cnblogs.com/yym2013