hdu 1785(对向量排序)

/*
*  题目要求：对一组向量按与x轴的正向夹角从小到大排序
*  解法：借用求凸包的排序规则即可 
*/

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

const int N = 105;
const double eps = 1e-8;

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

bool isZero(double x) {
    return (x > 0 ? x : -x) < eps;
}

double dis(point A, point B) {
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}

double crossProd(point A, point B, point C) {
    return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
}

int cmp(const void *a, const void *b) {//按x轴的正向夹角从小到大排序
    point *c = (point *)a;
    point *d = (point *)b;
    double k = crossProd(p[0], *c, *d);
    if (k<eps || (isZero(k)&&dis(p[0], *c)>dis(p[0], *d))) return 1;
    return -1;
}

void solve(int n) {
    qsort(p+1, n, sizeof(point), cmp);
    return ;
}

int main() {
    int n;
    p[0].x = p[0].y = 0;
    while (scanf("%d", &n) && n >= 1) {
        for (int i=1; i<=n; ++i) scanf ("%lf%lf", &p[i].x, &p[i].y);
        solve(n);
        printf ("%.1lf %.1lf", p[1].x, p[1].y);
        for (int i=2; i<=n; ++i) printf (" %.1lf %.1lf", p[i].x, p[i].y);
        printf ("\n");
    }
    return 0;
}