Triangle Partition
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 304 Accepted Submission(s): 170
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.
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.
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3*n个点,求你构造出的n个三角形的id位置(之前输入的位置)是哪些 ,三角形不能重合
题解:按照x轴排序,然后依次输出就行,注意只能用scanf输入,不然会T
#include <bits/stdc++.h> using namespace std; const int maxn = 1e4+5; struct node{ int x,y,id; bool operator < (const node & a)const{ return x<a.x; } }p[maxn]; int main(int argc, char const *argv[]) { int T; scanf("%d",&T); while(T--){ int n; scanf("%d",&n); for(int i=1;i<=n*3;i++){ scanf("%d %d",&p[i].x,&p[i].y); p[i].id=i; } sort(p+1,p+n*3+1); for(int i=1;i<=n*3;i+=3){ printf("%d %d %d ",p[i].id,p[i+1].id,p[i+2].id); } } return 0; }