HDU 6300 Triangle Partition (计算几何 + 点集划分)*_点集划分计算几何-优快云博客

本文介绍了一种解决特定几何问题的算法——三点匹配。该算法旨在利用输入的3n个不共线点，通过合理排序和分组，构造出n个互不相交的三角形。文章详细解释了算法的具体实现过程，包括如何对点进行排序以确保能够正确形成三角形。

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Triangle Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1370 Accepted Submission(s): 688
Special Judge

Problem Description

Chiaki has 3n points p1,p2,…,p3n . It is guaranteed that no three points are collinear.
Chiaki would like to construct n disjoint triangles where each vertex comes from the 3n points.

Input

There are multiple test cases. The first line of input contains an integer T , indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤1000 ) -- the number of triangle to construct.
Each of the next 3n lines contains two integers xi and yi (−109≤xi,yi≤109 ).
It is guaranteed that the sum of all n does not exceed 10000 .

Output

For each test case, output n lines contain three integers ai,bi,ci (1≤ai,bi,ci≤3n ) each denoting the indices of points the i -th triangle use. If there are multiple solutions, you can output any of them.

Sample Input

1 1 1 2 2 3 3 5

Sample Output

1 2 3

Source

2018 Multi-University Training Contest 1

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#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<queue>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define ll long long
#define maxn 3005
#define eps 0.0000001
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000") ///在c++中是防止暴栈用的
int t,n,mod=1e9+1;
bool Equal(double x,double y)
{
if(abs(x-y)<=eps) return true;
return false;
}
/*
给定3n个平面点，输出n组每组三个序号，
使序号点每次都能构成三角形（没有超过三个点共线）。

重点在如何对点集排序，即找出一条拟定的路线排序，
在平面上若干点中，如何画出一条路径线呢?

一开始我想到的是斜率，看了别人的思路才知道又想烦了。
其实只要按y轴，x轴划分即可。
从上往下，从左往右这样划分，
找不到反例有不符合题意的情况

*/
struct node
{
double x,y,xie;
int id;
node(){}
};

node seq[maxn];

bool cmp(node a,node b)
{
if(a.y==b.y) return a.x<b.x;
return a.y<b.y;///在多个点中拟定一条顺序线，使之符合题意
}

int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);

for(int i=1;i<=3*n;i++)
{
scanf("%lf%lf",&seq[i].x,&seq[i].y);
seq[i].id = i;
}
sort(seq+1,seq+3*n+1,cmp);
for(int i=1;i<=n;i++) printf("%d %d %d\n",seq[3*i-2].id,seq[3*i-1].id,seq[3*i].id);
}
return 0;
}