2019年杭州师范大学第十二届程序设计竞赛 K:Little Sub and Triangles(三角形面积公式+二分)-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41117236/article/details/88628122

博客围绕“Little Sub and Triangles”题目展开，给出已知三角形三个顶点坐标求面积公式S=0.5*fabs((x2 - x1)*(y3 - y1)-(x3 - x1)*(y2 - y1))，通过遍历求所有三角形面积，排序二分统计个数来解题，还给出题目描述、输入输出要求及示例。

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Little Sub and Triangles

【题解】

数学太差推错公式orz

已知三角形三个顶点坐标求三角形面积公式：S=0.5*fabs((x2-x1)*(y3-y1)-(x3-x1)*(y2-y1))

遍历求出所有三角形的面积，排序二分统计个数即可。

【代码】

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
double a[255][3];
double cul_s(int x,int y,int z)
{
    double x1=a[x][0],y1=a[x][1],x2=a[y][0],y2=a[y][1],x3=a[z][0],y3=a[z][1];
    double S=0.5*fabs((x2-x1)*(y3-y1)-(x3-x1)*(y2-y1));
    return S;
}
int main()
{
    int n,q; scanf("%d%d",&n,&q);
    for(int i=0;i<n;i++) scanf("%lf%lf",&a[i][0],&a[i][1]);
    vector <double> vec;
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++)
            for(int k=j+1;k<n;k++)
                vec.push_back(cul_s(i,j,k));
    sort(vec.begin(),vec.end());
    while(q--){
        double l,r; scanf("%lf%lf",&l,&r);
        int a=lower_bound(vec.begin(),vec.end(),l-eps)-vec.begin();
        int b=lower_bound(vec.begin(),vec.end(),r+eps)-vec.begin();
        printf("%d\n",b-a);
    }
    return 0;
}

【题目】

Little Sub and Triangles AC

Time Limit: 2 s Memory Limit: 256 MB

Submission：874 AC：94 Score：100.00

Description

Little Sub loves triangles. Now he has a problem for you.

Given n points on the two-dimensional plane, you have to answer many queries. Each query require you to calculate the number of triangles which are formed by three points in the given points set and their area S should satisfy l ≤ S ≤ r.

Specially, to simplify the calculation, even if the formed triangle degenerate to a line or a point which S = 0, we still consider it as a legal triangle.