HDU6300 Triangle Partition

Triangle Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 581    Accepted Submission(s): 303
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

 题意：

给你n个三角形，再给你3n个点，然后让你将所有点进行分配，让任意两个三角形不相交，而且在其中不会有三个点共线的情况。

算法：

说不上什么算法，你就是结构体排序一下，然后从左往右每三个就输出就行了，因为在其中不会三个点共线，所以最多两个共线，这样就保证了不会有线相交。

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;

struct point
{
    int x,y,id;
}p[3005];

bool cmp(point u,point v)
{
    if(u.x == v.x) return u.y > v.y;
    return u.x < v.x;
}

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].x,&p[i].y);
            p[i].id = i+1;
        }
        sort(p,p+(3*n),cmp);
        for(int i = 0 ; i < 3*n ; i += 3)
        {
            printf("%d %d %d\n",p[i].id,p[i+1].id,p[i+2].id);
        }
    }
    return 0;
}

补题才过，一开始没看这题，觉得这题更简单............自己太菜了。

菜得不一样，菜出新高度。