2018HDU多校训练一 C -Triangle Partition

最新推荐文章于 2020-06-15 20:49:10 发布

转载最新推荐文章于 2020-06-15 20:49:10 发布 · 147 阅读

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原文链接：http://www.cnblogs.com/songorz/p/9386492.html

本文解析了一道算法题目，任务是在给定的3n个不共线点中构造n个互不相交的三角形。文章提供了完整的C++代码实现，通过先读取输入的点坐标并排序，然后依次选取每三个点来构成一个三角形。

Chiaki has 3n3n points p1,p2,…,p3np1,p2,…,p3n. It is guaranteed that no three points are collinear.
Chiaki would like to construct nn disjoint triangles where each vertex comes from the 3n3n points.

Input

There are multiple test cases. The first line of input contains an integer TT, indicating the number of test cases. For each test case:
The first line contains an integer nn (1≤n≤10001≤n≤1000) -- the number of triangle to construct.
Each of the next 3n3n lines contains two integers xixi and yiyi (−109≤xi,yi≤109−109≤xi,yi≤109).
It is guaranteed that the sum of all nn does not exceed 1000010000.

Output

For each test case, output nn lines contain three integers ai,bi,ciai,bi,ci (1≤ai,bi,ci≤3n1≤ai,bi,ci≤3n) each denoting the indices of points the ii-th triangle use. If there are multiple solutions, you can output any of them.

Sample Input

Sample Output

1 2 3

#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
typedef long long ll;
typedef pair<int,int> PII;
const int N=101000;
int T,n,x,y;
pair<PII,int> p[N];
int main() 
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(int i=0;i<3*n;i++) 
		{
			scanf("%d%d",&x,&y);
			p[i]=mp(mp(x,y),i);
		}
		sort(p,p+3*n);
		for(int i=0;i<n;i++) 
			printf("%d %d %d\n",p[3*i].second+1,p[3*i+1].second+1,p[3*i+2].second+1);
	}
	return 0;
}

转载于:https://www.cnblogs.com/songorz/p/9386492.html