Triangle Partition

最新推荐文章于 2020-06-28 21:43:39 发布

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本文介绍了一个简单的算法，用于解决给定3n个不共线点时如何构造n个互不相交的三角形的问题。通过将这些点按x坐标排序并依次选择每三个连续的点来形成三角形，确保了所构造的三角形满足题目的要求。

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

思路：

题目说了任意三点都不相交，所以按照x坐标排序一下，从左往右输出点输入时的顺序就行。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

struct point{
    int x;
    int y;
    int pos;

}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].pos=i;
        }
        sort(a+1,a+3*n+1,cmp);
        for(int i=1;i<=3*n;++i)
            printf("%d ",a[i].pos);
        cout<<endl;
    }
    return 0;
}