Triangle Partition

最新推荐文章于 2020-06-28 21:43:39 发布

原创最新推荐文章于 2020-06-28 21:43:39 发布 · 266 阅读

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本文介绍了一道关于从给定点集中构造不相交三角形的算法题目，通过示例展示了如何利用C++进行实现。重点在于理解题目要求并正确输出各三角形顶点的索引。

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

代码如下：

#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
struct point
{
    int x;
    int y;
    int z;
}a[3005];
bool cmp(point a,point b)
{
    return a.x<b.x;
}
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",&a[i].x,&a[i].y);
            a[i].z=i;
        }
        sort(a+1,a+3*n+1,cmp);
        for(int i=1;i<=3*n;++i)
            printf("%d ",a[i].z);
        cout<<endl;
    }
    return 0;
}