2018 Multi-University Training Contest 1 1003 Triangle Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 2140    Accepted Submission(s): 925
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

一开始读错题意了orz，以为是要判断三点不共线输出全部可以组合的点，幸亏队友拨乱反正。

看明白题意很简单，因为不存在三点共线情况所以直接将x坐标排序就能生成不重合的几个三角形了。

#include <iostream>
#include <algorithm>
using namespace std;
struct Node 
{
	int x;
	int y;
	int num;
}node[3010];
bool cmp(const Node &a,const Node &b)
{
		return a.x<b.x;
}
int main()
{
	int T,n;
	int i;
	int cnt=0;
	scanf("%d",&T);
	
	while(T--)
	{
		scanf("%d",&n);
		for(i=1;i<=3*n;i++)
		{
			scanf("%d%d",&node[i].x,&node[i].y);
			node[i].num=i;
		}
		sort(node+1,node+3*n+1,cmp);
//		for(i=1;i<=3*n;i++)
//		{
//			printf("%d,%d %d\n",node[i].x,node[i].y,node[i].num);
//		}
		for(i=1;i<=3*n;i++)
		{
			if(i%3==0)
				printf("%d\n",node[i].num);
			else
				printf("%d ",node[i].num);
		}
	}
	return 0;
}