Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) C. Constellation

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于 2019-02-22 19:09:00 发布

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原文链接：http://www.cnblogs.com/-yjun/p/10420122.html

本文介绍了一种解决特定几何问题的方法，通过排序和检查点的相对位置，找到三个星点形成满足特定条件的三角形，确保所有其他点严格位于该三角形外。

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time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Cat Noku has obtained a map of the night sky. On this map, he found a constellation with n stars numbered from 1 to n. For each i, the i-th star is located at coordinates (xi, yi). No two stars are located at the same position.
In the evening Noku is going to take a look at the night sky. He would like to find three distinct stars and form a triangle. The triangle must have positive area. In addition, all other stars must lie strictly outside of this triangle. He is having trouble finding the answer and would like your help. Your job is to find the indices of three stars that would form a triangle that satisfies all the conditions.
It is guaranteed that there is no line such that all stars lie on that line. It can be proven that if the previous condition is satisfied, there exists a solution to this problem.
Input
The first line of the input contains a single integer n (3 ≤ n ≤ 100 000).
Each of the next n lines contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109).
It is guaranteed that no two stars lie at the same point, and there does not exist a line such that all stars lie on that line.
Output
Print three distinct integers on a single line — the indices of the three points that form a triangle that satisfies the conditions stated in the problem.
If there are multiple possible answers, you may print any of them.
Examples
Input
Copy
3
0 1
1 0
1 1
Output
Copy
1 2 3
Input
Copy
5
0 0
0 2
2 0
2 2
1 1
Output
Copy
1 3 5
Note
In the first sample, we can print the three indices in any order.
In the second sample, we have the following picture.

Note that the triangle formed by starts 1, 4 and 3 doesn't satisfy the conditions stated in the problem, as point 5 is not strictly outside of this triangle (it lies on it's border).

题解:这是一道几何题,比赛时不会写,嘿嘿.

思路:把点用结构数组保存起来,来和排序,先排x,x相等再排y,递增排序(因为排序后原来的顺序会打乱,所以要在输入时保存原来的位置).

例如:

0 0
0 2
2 0
2 2
1 1

排完序后为:

0 0
0 2
1 1
2 0
2 2

因为是排完序后的,所以第一个点和第二个点的连线上不会有其他的点.

所以我们只要在剩下的点找一个点,只要不和这两个点在同一条直线上(即不共线),

判断两条线段是否共线可以用向量来判断.例如向量a(x1,y1),向量b为(x2,y2),当x1y2==x2y1时.说明向量共线.共线就不符合.

#include <bits/stdc++.h>
typedef long long ll;
const int N=1e5+5;
using namespace std;
struct point{
    int x,y,po;
}p[N];
bool comp(point a,point b){
    if(a.x<b.x) return true;
    if(a.x>b.x) return false;
    if(a.y<b.y) return true;
    else return false;
}
bool check(int n){
    ll x1=p[0].x-p[1].x;
    ll y1=p[0].y-p[1].y;
    ll x2=p[n].x-p[1].x;
    ll y2=p[n].y-p[1].y;
    if(x1*y2-x2*y1==0) return false;
    return true;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d%d",&p[i].x,&p[i].y),p[i].po=i+1;
    sort(p,p+n,comp);
    //for(int i=0;i<n;i++) printf("%d %d\n",p[i].x,p[i].y);
    //cout << "Hello world!" << endl;
    for(int i=2;i<n;i++)
        if(check(i)){
            printf("%d %d %d\n",p[0].po,p[1].po,p[i].po);
            break;
        }
    return 0;
}

转载于:https://www.cnblogs.com/-yjun/p/10420122.html