bzoj 1132 [POI2008]Tro 几何

[POI2008]Tro

Time Limit: 20 Sec  Memory Limit: 162 MB
Submit: 1796  Solved: 604
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Description

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

Input

第一行给出数字N,N在[3,3000] 下面N行给出N个点的坐标,其值在[0,10000]

Output

保留一位小数,误差不超过0.1

Sample Input

5
0 0
1 2
0 2
1 0
1 1

Sample Output

7.0

HINT

枚举起点，然后求出以该点为起点的所有向量，然后求面积就可以了。

 #include<cstring>
 #include<cmath>
 #include<cstdio>
 #include<iostream>
 #include<algorithm>

 #define N 3007
 #define ll long long
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 int n;ll ans;
 struct Node
 {
     int x,y;
     friend inline ll operator*(Node x,Node y)
     {
         return x.x*y.y-x.y*y.x;
     }
     friend inline bool operator<(Node x,Node y)
     {
         if (x.y==y.y) return x.x<y.x;
         return x.y<y.y;
     }
 }a[N],b[N];
 bool cmp(Node x,Node y)
 {
     return x*y>;
 }

 void solve()
 {
     sort(a+,a+n+);
     for (int i=;i<=n-;i++)
     {
         int tot=;ll sumx=,sumy=;
         for (int j=i+;j<=n;j++)
             b[++tot].x=a[j].x-a[i].x,
             b[tot].y=a[j].y-a[i].y;
         sort(b+,b+tot+,cmp);
         for (int j=;j<=tot;j++)
             sumx+=b[j].x,
             sumy+=b[j].y;
         for (int j=;j<=tot;j++)
         {
             sumx-=b[j].x;
             sumy-=b[j].y;
             ans+=(ll)b[j].x*sumy-b[j].y*sumx;
         }
     }
 }
 int main()
 {
     n=read();
     for (int i=;i<=n;i++)
         a[i].x=read(),a[i].y=read();
     solve();
     if (ans&) printf("%lld.5",ans/);
     else printf("%lld.0",ans/);
 }
 #undef ll