BZOJ 1132 Tro

Tro

【问题描述】

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

【输入格式】

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]

【输出格式】

保留一位小数,误差不超过0.1

【样例输入】

5
0 0
1 2
0 2
1 0
1 1

【样例输出】

7.0

---

题解:

叉积之和

我们以每个点为原点，维护前缀和

为了保证夹角不超过π，先按水平序排序

为了保证面积都是正值，按极角序排序

 #include<cmath>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 struct point
 {
     long long x, y;
     friend inline long long operator * (point a, point b)
     {
         return a.x * b.y - a.y * b.x;
     }
     friend inline point operator - (point a, point b)
     {
         return (point) {a.x - b.x, a.y - b.y};
     }
     inline void empty()
     {
         x = y = ;
     }
 };
 point operator + (point a, point b)
 {
     point c;
     c.x = a.x + b.x;
     c.y = b.y + b.y;
     return c;
 }
 const int maxn = ;
 int n;
 point p[maxn];
 inline bool lev(point a, point b)
 {
     if(a.y != b.y) return a.y < b.y;
     return a.x < b.x;
 }

 int num;
 point np[maxn];
 inline bool ang(point a, point b)
 {
     return a * b > ;
 }
 point sum;
 long long ans;
 int main()
 {
     scanf("%d", &n);
     for(int i = ; i <= n; ++i) scanf("%lld %lld", &p[i].x, &p[i].y);
     sort(p + , p +  + n, lev);
     for(int k = ; k <= n; ++k)
     {
         num = ;
         for(int i = k + ; i <= n; ++i) np[++num] = p[i] - p[k];
         sort(np + , np +  + num, ang);
         sum.empty();
         for(int i = ; i <= num; ++i)
         {
             ans += sum * np[i];
             sum.x += np[i].x;
             sum.y += np[i].y;
         }
     }
     printf("%lld.%d", ans >> , (ans & ) ?  : );
 }