【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+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int dx[]={-,,,};
       int dy[]={,,-,};

 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>;
 }

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

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

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