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 = 0;
    }
};
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 = 3333;
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 > 0;
}
point sum;
long long ans;
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%lld %lld", &p[i].x, &p[i].y);
    sort(p + 1, p + 1 + n, lev);
    for(int k = 1; k <= n; ++k)
    {
        num = 0;
        for(int i = k + 1; i <= n; ++i) np[++num] = p[i] - p[k];
        sort(np + 1, np + 1 + num, ang);
        sum.empty();
        for(int i = 1; i <= num; ++i)
        {
            ans += sum * np[i];
            sum.x += np[i].x;
            sum.y += np[i].y;
        }
    }
    printf("%lld.%d", ans >> 1, (ans & 1) ? 5 : 0);
}