HDU 2018 Multi-University Training Contest 1 Triangle Partition 【YY】

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6300

Triangle Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 2964    Accepted Submission(s): 1474
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

 

Source

2018 Multi-University Training Contest 1

 

题意概括：

给出 3*N 个点的坐标（保证点不共线）；

用这 3*N 个点构造出 N 个不相交的三角形；

每一行输出一个三角形的坐标编号。

解题思路：
因为点都不共线，所以按横坐标排个序，三个三个点构造的三角形不相交。

AC code：

#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAXN = 3e3+10;

struct data
{
    int x, y;
    int no;
}node[MAXN];

bool cmp(data a, data b)
{
    return a.x < b.x;
}

int main()
{
    int N;
    int T_case;
    while(~scanf("%d", &T_case)){
        while(T_case--){
            scanf("%d", &N);
            int maxx = 3*N;
            for(int i = 0; i < maxx; i++){
                scanf("%d %d", &node[i].x, &node[i].y);
                node[i].no = i+1;
            }
            sort(node, node+maxx, cmp);

            for(int i = 0; i+2 < maxx; i+=3){
                printf("%d %d %d
", node[i].no, node[i+1].no, node[i+2].no);
            }
        }
    }
    return 0;
}