【BZOJ1132】Tro（叉积）

题意：平面上有N个点. 求出所有以这N个点为顶点的三角形的面积和

N<=3000 N个点的坐标,其值在[0,10000]

思路：按从左到右的预处理点排序

每次枚举最左点作为原点，把叉积从大到小排序

面积用叉积算，因为每次以最左的点作为原点，叉积一定都大于0

2S=xi*yj-yi*xj，xi和yi已经固定，只要维护xj和yj的后缀和就好

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
//typedef pair<ll,ll>P;
#define N  100100
#define M  2000010
#define fi first
#define se second
#define MP make_pair
#define pb push_back
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=1e9+7,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};


struct P
{
    ll x,y;
}p[N],t[N];

int n;

ll operator*(P a,P b)
{
    return a.x*b.y-a.y*b.x;
}

ll operator<(P a,P b)
{
    return a.y<b.y||(a.y==b.y&&a.x<b.x);
}

bool cmp(P a,P b)
{
    return a*b>0;
}

ll read()
{
   ll v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}


void solve()
{
    sort(p+1,p+n+1);
    ll ans=0;
    rep(i,1,n-2)
    {
        int m=0;
        ll sx=0,sy=0;
        rep(j,i+1,n)
        {
            m++;
            t[m].x=p[j].x-p[i].x;
            t[m].y=p[j].y-p[i].y;
        }
        sort(t+1,t+m+1,cmp);
        rep(j,1,m)
        {
            sx+=t[j].x;
            sy+=t[j].y;
        }
        rep(j,1,m)
        {
            sx-=t[j].x;
            sy-=t[j].y;
            ans+=t[j].x*sy-t[j].y*sx;
        }
    }
    if(ans%2==1) printf("%lld.5
",ans/2);
     else printf("%lld.0
",ans/2);
}

int main()
{
    n=read();
    rep(i,1,n) p[i].x=read(),p[i].y=read();
    solve();
    return 0;
}