CodeForces - 13D Triangles

本文介绍了一种使用向量叉积来判断一个点是否位于三角形内的算法实现。通过分析向量积的方向性,该方法能够有效地确定点的位置,并通过实例代码展示了如何利用这一原理进行计算。

本题是向量叉积的应用,应用向量叉积可以判断一个点是否在三角形内部。

向量积 axb =|a|*|b|*sin<a,b>

当b在a的逆时针方向,向量积是正数,反之,是负数。


由图可知三角形内部的点都在三角形边的同一侧。

可以结合此性质解决本题。


绿色和红色区域可以根据三角形行走的方向相消为0


#include<iostream>
#include<cstring>
#include<cmath>
#include<assert.h>
#include<stdio.h>
using namespace std;

typedef __int64 lld;
const int INF=1000000000;
struct P
{
    lld x,y;
    P(){};
    P(lld x,lld y):x(x),y(y){};
    P operator -(const P & b)const
    {
        P tmp(x-b.x,y-b.y);
        return tmp;
    }
};

lld cross(P a,P b)
{
    return a.x*b.y-a.y*b.x;
}

P red[505],blue[505];
int cnt[505][505];
int main()
{
    int i,j,k;
    int N,M;
    lld x,y;
    memset(cnt,0,sizeof(cnt));
    cin>>N>>M;
    assert(N>=0&&N<=500);
    assert(M>=0&&M<=500);
    for(i=0;i<N;i++)
    {
        cin>>x>>y;
        assert(x>=-INF&&x<=INF);
        assert(y>=-INF&&y<=INF);
        red[i]=P(x,y);
    }
    for(i=0;i<M;i++)
    {
        cin>>x>>y;
        assert(x>=-INF&&x<=INF);
        assert(y>=-INF&&y<=INF);
        blue[i]=P(x,y);
    }

    for(i=0;i<N;i++)
    {
        for(j=0;j<N;j++)
        {
            if(i==j||red[i].x>red[j].x)continue;
            for(k=0;k<M;k++)
            {
                if(blue[k].x>=red[i].x&&blue[k].x<red[j].x&&cross(blue[k]-red[i],red[j]-red[i])>0)
                    cnt[i][j]++;
            }
            cnt[j][i]=-cnt[i][j];
        }
    }

    int ans=0;
    for(i=0;i<N;i++)
    {
        for(j=i+1;j<N;j++)
            for(k=j+1;k<N;k++)
        {
            if(cnt[i][j]+cnt[j][k]+cnt[k][i]==0)
                ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}


### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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