POJ 3348 Cows 凸包面积

题目大意：连接个点，求围成的面积并除以50（向下取整）

题目思路：就是求凸包面积：按逆时针方向为凸包上每条边指定方向，对于每条边AB，累加（AXB）/2的值。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define MAX 100005

using namespace std;

struct node
{
    int x,y,id;
}point[MAX];

int n,ans[MAX],Stuck[MAX],top;

bool cmp(struct node A,struct node B)
{
    if(A.x < B.x)
        return true;
    else if(A.x==B.x && A.y < A.y)
        return true;
    return false;
}

int Cross(int x1,int y1,int x2,int y2,int x3,int y3)
{
    return (x1-x2)*(y1-y3)-(x1-x3)*(y1-y2);
}

double Area()
{
    double sum=0.0;
    for(int i=1;i<=top;i++)
        sum+=(point[Stuck[i-1]].x*point[Stuck[i]].y-point[Stuck[i]].x*point[Stuck[i-1]].y);
    return fabs(sum/2);
}

void Graham()
{
    int len;
    top=1;
    for(int i=0;i<2;i++)
        Stuck[i]=i;
    for(int i=2;i<n;i++)
    {
        while(top>0 && Cross(point[Stuck[top]].x,point[Stuck[top]].y,point[Stuck[top-1]].x,point[Stuck[top-1]].y,point[i].x,point[i].y)<=0)
            top--;
        Stuck[++top]=i;
    }
    len=top;
    Stuck[++top]=n-2;
    for(int i=n-3;i>=0;i--)
    {
        while(top!=len && Cross(point[Stuck[top]].x,point[Stuck[top]].y,point[Stuck[top-1]].x,point[Stuck[top-1]].y,point[i].x,point[i].y)<=0)
            top--;
        Stuck[++top]=i;
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(ans,0,sizeof(ans));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&point[i].x,&point[i].y);
            point[i].id=i;
        }
        if(n<3)
        {
            printf("0\n");
            continue;
        }
        sort(point,point+n,cmp);
        Graham();
        double ans=Area();
        int sum=ans/50;
        printf("%d\n",sum);
    }
    return 0;
}