Triangle Partition

最新推荐文章于 2019-04-18 11:33:48 发布

漂流瓶终结者

最新推荐文章于 2019-04-18 11:33:48 发布

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文章标签：思维

本文链接：https://blog.youkuaiyun.com/ZCMU_2024/article/details/81335076

本文介绍了一个简单的算法，用于从一组不共线的点中构造出指定数量的互不相交的三角形。通过将点按横坐标排序并依次选取每三个连续的点形成三角形，实现了题目的要求。

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Descrition

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 (−10^9≤xi,yi≤10^9).
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

解析

题目关键句是任意三点不共线，那么很简单，只要将坐标按横坐标从小到大排序，连续3个点组成一个三角形即可

代码

#include<stdio.h>
#include<algorithm>
using namespace std;
#define MAX 40005
struct point{
 int xi;
 int yi;
 int index;
}p[MAX];
bool cmp(point a,point b)
{
    if(a.xi==b.xi)
        return a.yi<b.yi;
    return a.xi<b.xi;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<3*n;i++)
        {
            scanf("%d%d",&p[i].xi,&p[i].yi);
            p[i].index=i+1;
        }
        sort(p,p+3*n,cmp);
        for(int i=0;i<n;i++)
        {
            printf("%d %d %d\n",p[i*3].index,p[i*3+1].index,p[i*3+2].index);
        }
    }
    return 0;
}