1003. Triangle Partition

• Time limit: 1 second
• Memory limit: 32 megabytes
  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3n个任意三点不在同一直线上的点，找出n个不相交的三角形
标程思路：求个凸包，然后选择凸包一条边 ，然后找个和 夹角最小的点 ，把 当做一个三角形删掉即可。 这样做个n次，显然这样求出来的三角形们是合法的。

但实际上，满足不相交只要按照从左到右，从上到下的顺序删去点即可

标程：(凸包的代码我还没手动敲过 mark)

#include <cstdio>
#include <vector>

using int64 = long long;

struct point {
  int64 x, y;
  int i;
  point(int64 x = 0, int64 y = 0): x(x), y(y) {}
  point operator + (const point& rhs) const {
    return {x + rhs.x, y + rhs.y};
  }
  point operator - (const point& rhs) const {
    return {x - rhs.x, y - rhs.y};
  }
  int64 det(const point& rhs) const {
    return x * rhs.y - y * rhs.x;
  }
  int64 dot(const point& rhs) const {
    return x * rhs.x + y * rhs.y;
  }
};

const int N = 5e3 + 10;

point p[N];

int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) {
    int n;
    scanf("%d", &n);
    n *= 3;
    for (int i = 0; i < n; ++i) {
      scanf("%lld%lld", &p[i].x, &p[i].y);
      p[i].i = i + 1;
    }
    for (int it = 0; it < n / 3; ++it) {
      int m = n - it * 3;
      int a = 0, b = -1, c = -1;
      // lowest point p_a
      for (int i = 1; i < m; ++i) {
        if (p[i].y < p[a].y || (p[i].y == p[a].y && p[i].x < p[a].x)) a = i;
      }
      // right most point p_b
      for (int i = 0; i < m; ++i) if (i != a) {
        if (b == -1 || (p[b] - p[a]).det(p[i] - p[a]) < 0) b = i;
      }
      // second right most point p_c
      for (int i = 0; i < m; ++i) if (i != a && i != b) {
        if (c == -1 || (p[c] - p[a]).det(p[i] - p[a]) < 0) c = i;
      }
      printf("%d %d %d\n", p[a].i, p[b].i, p[c].i);
      for (int i = 0, j = 0; i < m; ++i) if (i != a && i != b && i != c) {
        p[j++] = p[i];
      }
    }
  }
  return 0;
}

比赛代码：

//B17040312
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

struct point{
    int x;
    int y;
}p[30005];

int pos[30005];

bool cmp(int a , int b){
    if(p[a].x<p[b].x) return true;
    else if(p[a].x>p[b].x) return false;
    else{
        return p[a].y<p[b].y;
    }
}

int main(){
    int t,n;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        for(int i = 0; i < 3*n; i++){
            scanf("%d %d", &p[i].x,&p[i].y);
            pos[i] = i;
        }
        sort(pos, pos+3*n, cmp);
        for(int i = 0; i < 3*n; i++){
            printf("%d", pos[i]+1);
            if((i+1)%3==0&&i!=3*n-1) printf("\n");
            else printf(" ");
        }
    }
    return 0;
}