Triangle Partition水题

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本文详细解析了如何解决三角形划分问题，通过确保无三点共线的条件下，将3n个点分配为n个互不相交的三角形。文章提供了具体算法思路，即先按x坐标对点进行排序，然后依次选取点构成三角形，实现了解决方案的构建。代码示例使用C++实现，展示了完整的输入输出流程。

Triangle Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 891 Accepted Submission(s): 463
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 2

2 3

3 5

Sample Output

1 2 3

Source

2018 Multi-University Training Contest 1

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这道题主要是想明白由于给的所有点都不共线，就说明一条线上最多有两个点。所以将其以 x 从小到大进行排序，然后直接输出就可以了。

代码：

#include<bits/stdc++.h>
using namespace std;
struct node
{
    long long x,y;
    int id;
}lala[10010];
int cmp(node a,node b)
{
    return a.x<b.x;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=3*n;i++)
        {
            scanf("%lld%lld",&lala[i].x,&lala[i].y);
            lala[i].id=i;
        }
        sort(lala+1,lala+1+3*n,cmp);
        for(int i=1;i<=3*n;i++)
        {
            if(i%3==1||i%3==2)
                cout<<lala[i].id<<" ";
           
            else if(i%3==0)
                cout<<lala[i].id<<endl;
        }
    }
    return 0;
}