Ural_1207. Median on the Plane(计算几何)

　　/*思路完全搞乱，开始就没想清楚就写。我晕，各种WA。

思路：

　　找到所有点中最下边一层点里边最靠左的点d。然后求d到其他每个点连线与x轴的
夹角Θ（0 <= Θ <= π 因为d的纵坐标最小）。然后从小到大排序，找到存角度的数组
里n/2号点就是要找的另一个点。*/

My Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define pi 3.1415926535

using namespace std;

const int N = 10010;

struct point{
    double x;
    double y;
}p[N];

struct angle{
    double ang;
    int i;
}a[N];

point tmp;

bool cmp(angle a, angle b){
    return a.ang < b.ang;
}
int main(){
    //freopen("data.in", "r", stdin);

    int n, i, d, j;
    double b, c;

    scanf("%d", &n);
    scanf("%lf%lf", &p[1].x, &p[1].y);
    b = p[1].x; c = p[1].y; d = 1;

    for(i = 2; i <= n; i++){
        scanf("%lf%lf", &p[i].x, &p[i].y);
        if(c > p[i].y){
            b = p[i].x; c = p[i].y; d = i;
        }
        if(c == p[i].y && b > p[i].x){
            b = p[i].x; c = p[i].y; d = i;
        }
    }

    for(j = 1, i = 1; i <= n; i++){
        if(i != d){
            a[j].i = i;
            b = p[i].x - p[d].x;
            c = p[i].y - p[d].y;
            if(b == 0)    {a[j++].ang = pi/2; continue;}
            if(c == 0)    {a[j++].ang = 0; continue;}
            if(p[i].x < p[d].x)
                a[j++].ang = pi - atan(c/b);
            else
                a[j++].ang = atan(c/b);
        }
    }
    sort(a+1, a+n, cmp);
    /*for(i = 1; i < n; i++){
        printf("%lf %d\n", a[i].ang, a[i].i);
    }*/
    printf("%d %d\n", d, a[n/2].i);
    return 0;
}