Ural 1207. Median on the Plane（计算几何）

http://acm.timus.ru/problem.aspx?num=1207 

题意：给你n个点，让你在n个点当中选两个点，使得两边点的数量相等；

题解：想找出最靠左边的最下边的点，以这个点求极角，然后排序 ，它的一半就是我们所要求的点

code：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
#define eps 1e-8
//点
struct POINT
{
double x, y;
int pnd;
POINT(){ }
POINT(double a, double b){
x = a;
y = b;
}
}p[10005];
//线段
struct Seg
{
POINT a, b;
Seg() { }
Seg(POINT x, POINT y){
a = x;
b = y;
}
};
//叉乘
double cross(POINT o, POINT a, POINT b)
{
return (a.x - o.x) * (b.y - o.y) - (b.x - o.x) * (a.y - o.y);
}
//判断点在线段上
//bool On_Seg(POINT a, Seg s)
//{
// double maxx = max(s.a.x, s.b.x), minx = min(s.a.x, s.b.x);
// double maxy = max(s.a.y, s.b.y), miny = min(s.a.y, s.b.y);
// if(a.x >= minx && a.x <= maxx && a.y >= miny && a.y <= maxy) return true;
// return false;
//}
////判断线段相交
//bool Seg_cross(Seg s1, Seg s2)
//{
// double cs1 = cross(s1.a, s2.a, s2.b);
// double cs2 = cross(s1.b, s2.a, s2.b);
// double cs3 = cross(s2.a, s1.a, s1.b);
// double cs4 = cross(s2.b, s1.a, s1.b);
// // 互相跨立
// if(cs1 * cs2 < 0 && cs3 * cs4 < 0) return true;
// if(cs1 == 0 && On_Seg(s1.a, s2)) return true;
// if(cs2 == 0 && On_Seg(s1.b, s2)) return true;
// if(cs3 == 0 && On_Seg(s2.a, s1)) return true;
// if(cs4 == 0 && On_Seg(s2.b, s1)) return true;
// return false;
//}
////求两条线段的交点，但是，必须保证线段相交且不共线
////共线的话需要特判
//POINT Inter(Seg s1, Seg s2)
//{
// double k = fabs(cross(s1.a, s2.a, s2.b)) / fabs(cross(s1.b, s2.a, s2.b));
// return POINT((s1.a.x + s1.b.x * k) / (1 + k), (s1.a.y + s1.b.y * k) / (1 + k));
//}
////多边形面积，需要有顺序，顺(逆)时针。

//double area()
//{
// double ans = 0;
// for(int i = 1; i < top; i ++){
// ans += cross(p[0], p[i], p[i + 1]);
// }
// return ans;
//}
////找凸包基点排序
//bool cmp0(POINT a, POINT b)
//{
// if(a.y < b.y) return true;
// else if(a.y == b.y && a.x < b.x) return true;
// return false;
//}
//极角排序
double dis(POINT a , POINT b){
return sqrt( (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y) );
}
bool cmp1(POINT a, POINT b)
{
if(cross(p[0], a, b) > eps) return true;
else if(fabs(cross(p[0], a, b)) < eps && dis(p[0], a) - dis(p[0], b) > eps) return true;
return false;
}
////Graham_scan 求凸包.所求为纯净凸包...
//void Graham_scan()
//{
// sort(p, p + n, cmp0);
// sort(p + 1, p + n, cmp1);
// top = 0;
// p
= p[0];
// st[top ++] = p[0]; st[top ++] = p[1];
// for(int i = 2; i <= n; i ++){
// while(top > 2 && (cross(st[top - 1], st[top - 2], p[i]) > eps || fabs(cross(st[top - 1], st[top - 2], p[i])) < eps)) top --;
// st[top ++] = p[i];
// }
// top --;
//}
int main(){
int n;
while(cin >> n){
for(int i = 0;i < n;i++){
cin >> p[i].x >> p[i].y;
p[i].pnd = i + 1;
}
int tmp = 0;
for(int i = 1;i < n;i++)
{
if(p[i].x < p[tmp].x || (p[i].x == p[tmp].x && p[i].y < p[tmp].y))
tmp = i;
}
swap(p[0],p[tmp]);
sort(p+1,p+n,cmp1);
cout << p[0].pnd << " " << p[n/2].pnd << endl;;
}
}