Description

平面上有N個點. 求出所有以這N個點為頂點的三角形的面積和 N<=3000

Input

第一行給出數字N,N在[3,3000] 下面N行給出N個點的座標,其值在[0,10000]

Output

保留一位小數,誤差不超過0.1

Sample Input

5
0 0
1 2
0 2
1 0
1 1

Sample Output

7.0

Solution

\(ans=\frac{1}{2}\sum_{i=1}^n\sum_{j=i+1}^n\sum_{k=j+1}^n|(y_j-y_i)(x_k-x_i)-(y_k-y_i)(x_j-x_i)|\)
列舉第一個點，求出其它點的相對座標
然後為了去絕對值，讓所有點按計較排序，保證叉積是正的
\(ans_i=\frac{1}{2}\sum_{j=i+1}^n\sum_{k=j+1}^ny_j*x_k-y_k*x_j\)

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define REP(a,b,c) for(register int a=(b),a##end=(c);a<=a##end;++a)
#define DEP(a,b,c) for(register int a=(b),a##end=(c);a>=a##end;--a)
const int MAXN=3000+10;
int n,cnt;
ld ans;
struct point{
    int x,y;
    inline bool operator < (const point &A) const {
        return y<A.y;
    };
};
point pt[MAXN];
struct cross{
    int x,y;
    ld k;
    inline bool operator < (const cross &A) const {
        return k>A.k;
    };
};
cross cs[MAXN];
template<typename T> inline void read(T &x)
{
    T data=0,w=1;
    char ch=0;
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')w=-1,ch=getchar();
    while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
    x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
    if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
int main()
{
    read(n);
    REP(i,1,n)read(pt[i].x),read(pt[i].y);
    std::sort(pt+1,pt+n+1);
    REP(i,3,n)
    {
        REP(j,1,i-1)cs[j]=(cross){pt[i].x-pt[j].x,pt[i].y-pt[j].y,atan2((ld)(pt[i].x-pt[j].x),(ld)(pt[i].y-pt[j].y))};
        std::sort(cs+1,cs+i);ld sx=0,sy=0;
        REP(j,1,i-1)
        {
            if(j!=1)ans+=sx*(ld)cs[j].y-(ld)cs[j].x*sy;
            sx+=(ld)cs[j].x,sy+=(ld)cs[j].y;
        }
    }
    printf("%.1Lf\n",ans/2);
    return 0;
}