Little Sub and Triangles

最新推荐文章于 2022-09-18 20:14:22 发布

原创最新推荐文章于 2022-09-18 20:14:22 发布 · 164 阅读

0 ·

CC 4.0 BY-SA版权

省赛准备专栏收录该内容

62 篇文章

订阅专栏

博客围绕二维平面上n个点的三角形面积查询问题展开。给定n个点坐标及q个查询，每个查询需计算面积在指定区间[l, r]内的三角形数量。作者分享收获，包括用坐标求三角形面积公式、for循环控制输出及二分判断区间内个数等细节。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述
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.

输入
The fi rst line contains two integer n, q(1≤n≤250, 1≤q≤100000), indicating the total number of points.
All points will be described in the following n lines by giving two integers x, y(-107≤x, y≤107) as their coordinates.
All queries will be described in the following q lines by giving two integers l, r(0≤l≤r≤1018).

输出
Output the answer in one line for each query.

样例输入
复制样例数据
4 2
0 1
100 100
0 0
1 0
0 50
0 2
样例输出
3
1
这个题我主要收获了一个三角形的用坐标求面积的公式向量a（a,b）和向量b(x,y)的坐标方向相乘再相减，其实就是bx-ay，还有就是for循环控制输出，看看下面的for循环输出的其实形成三角形的数量这个对我来说其实就是这些细节总是把握不好，这次学会了从小到大进行控制。最后二分进行判断在区间里面的个数，注意sort一下

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
#define ll long long
struct point
{
    ll x,y;
}a[900];
ll b[9000000];
int main()
{
    int n,num=0,q;
    ll ans;
    scanf("%d %d",&n,&q);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld %lld",&a[i].x,&a[i].y);
         for(int j=1;j<i;j++){
            for(int k=j+1;k<i;k++){
                ans=abs((a[i].x-a[j].x)*(a[i].y-a[k].y)-(a[i].x-a[k].x)*(a[i].y-a[j].y));
                b[num++]=abs(ans);
            }
        }
    }
    sort(b,b+num);
    for(int i=0;i<q;i++)
    {
        ll l,r;
        scanf("%lld %lld",&l,&r);
        ll count1=(upper_bound(b,b+num,r*2)-b)-(lower_bound(b,b+num,l*2)-b);
        printf("%lld\n",count1);
    }
    return 0;
}