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
题意：给出3n个点，求一种方案，其中连成n个三角形，要求每个三角形不相交。（保证没有三个点在一条直线上）
思路：按照优先x其次y最小的排序，然后依次输出三个点。因为保证了三点不共线，这样排序x与y就保证了所有三角不相交。
这道题也不能想复杂了。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 3e3+10;
struct poin
{
int x,y,id;
}G[maxn];
int cmp(poin a,poin b)
{
if(a.x<b.x)return 1;
else if(a.x==b.x)
{
if(a.y<b.y)return 1;
return 0;
}
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
for(int i=1;i<=3*n;i++)
{
scanf("%d%d",&G[i].x,&G[i].y);
G[i].id=i;
}
sort(G+1,G+1+3*n,cmp);
for(int i=1;i<=3*n;i+=3)
{
printf("%d %d %d\n",G[i].id,G[i+1].id,G[i+2].id);
}
}
return 0;
}