poj3348（凸包面积模板）

思路：模板题，求面积就是以p0为统一端点，转换成求一些三角形的面积和。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod int(1e9+7)
#define pb push_back
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
#define eps 1e-8
int n,top,m,cnt;
double x,y;
const int N=1008;
struct point
{
    double x,y;
    point(){}
    point(double a,double b):x(a),y(b){}
    point operator-(point a){
        return point(x-a.x,y-a.y);
    }
    point operator+(point a){
        return point(x+a.x,y+a.y);
    }
    double operator*(point a){
        return x*a.y-y*a.x;
    }
    bool operator<(const point a) const{
        if(fabs(x-a.x)<eps) return y<a.y;
        return x<a.x;
    }
    double len(){
        return sqrt(x*x+y*y);
    }
}a[N],p[N];
double cj(point a,point b,point c) {return (a-c)*(b-c);}
void graham()
{
    sort(a+1,a+cnt+1);
    p[1]=a[1],p[2]=a[2],top=2;
    for(int i=3;i<=cnt;i++)
    {
        while(top>1&&cj(a[i],p[top],p[top-1])>=0) top--;
        p[++top]=a[i];
    }
    m=top;
    p[++top]=a[cnt-1];
    for(int i=cnt-2;i>=1;i--)
    {
        while(top>m&&cj(a[i],p[top],p[top-1])>=0) top--;
        p[++top]=a[i];
    }
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    si(n);
    FOR(i,1,n) sd(a[i].x),sd(a[i].y);
    sort(a+1,a+n+1);
    cnt=1;
    FOR(i,2,n)
    {
        if(fabs(a[i].x-a[cnt].x)>eps||fabs(a[i].y-a[cnt].y)>eps) a[++cnt]=a[i];
    }
    graham();
    double ans=0.0;
    int s;
    FOR(i,2,top-1) ans+=cj(p[i],p[i+1],p[1]);
    ans/=100.0;
    s=(int)ans;
    cout<<s<<endl;
    return 0;
}
poj3348（凸包面积模板）

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