凸包

/* 
 * 求凸包，Graham算法 
 * 点的编号0~n-1 
 * 返回凸包结果Stack[0~top-1]为凸包的编号 
 */ 
const int MAXN = 1010; 
Point list[MAXN]; 
int Stack[MAXN],top; 
//相对于list[0]的极角排序 
bool _cmp(Point p1,Point p2) 
{ 
	double tmp = (p1-list[0])^(p2-list[0]); 
  	if(sgn(tmp) > 0)return true; 
  	else if(sgn(tmp) == 0 && sgn(dist(p1,list[0]) - dist(p2,list[0])) <= 0) 
    	return true; 
  	else return false; 
} 
void Graham(int n) 
{ 
  	Point p0; 
  	int k = 0; 
  	p0 = list[0]; 
  	//找最下边的一个点 
  	for(int i = 1;i < n;i++) 
  	{ 
    	if( (p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x) ) 
    	{ 
      	p0 = list[i]; 
      	k = i; 
    	} 
  	} 
  	swap(list[k],list[0]); 
  	sort(list+1,list+n,_cmp); 
  	if(n == 1) 
  	{
	  top = 1; 
      Stack[0] = 0; 
      return; 
  	} 
  	if(n == 2) 
  	{ 
      top = 2; 
      Stack[0] = 0; 
      Stack[1] = 1; 
      return ; 
    } 
  	Stack[0] = 0; 
  	Stack[1] = 1; 
 	top = 2; 
  	for(int i = 2;i < n;i++) 
  	{ 
    	while(top>1 && sgn((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]]))<=0) 
      	top--; 
	  	Stack[top++] = i; 
  	} 
} 


3、平面最近点对（HDU 1007） 
#include <stdio.h> 
#include <string.h> 
#include <algorithm> 
#include <iostream> 
#include <math.h> 
using namespace std; 
const double eps = 1e-6; 
const int MAXN = 100010; 
const double INF = 1e20; 
struct Point 
{ 
    double x,y; 
}; 
double dist(Point a,Point b) 
{ 
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); 
} 
Point p[MAXN]; 
Point tmpt[MAXN]; 
bool cmpxy(Point a,Point b) 
{ 
    if(a.x != b.x)return a.x < b.x; 
    else return a.y < b.y; 
} 
bool cmpy(Point a,Point b) 
{ 
    return a.y < b.y; 
} 
double Closest_Pair(int left,int right) 
{ 
    double d = INF; 
    if(left == right)return d; 
    if(left + 1 == right) 
        return dist(p[left],p[right]); 
    int mid = (left+right)/2; 
    double d1 = Closest_Pair(left,mid); 
    double d2 = Closest_Pair(mid+1,right); 
    d = min(d1,d2); 
    int k = 0; 
    for(int i = left;i <= right;i++) 
    { 
        if(fabs(p[mid].x - p[i].x) <= d) 
            tmpt[k++] = p[i]; 
    } 
    sort(tmpt,tmpt+k,cmpy); 
    for(int i = 0;i <k;i++) 
    { 
        for(int j = i+1;j < k && tmpt[j].y - tmpt[i].y < d;j++) 
        { 
            d = min(d,dist(tmpt[i],tmpt[j])); 
        } 
    } 
    return d; 
} 
int main() 
{ 
    int n; 
    while(scanf("%d",&n)==1 && n) 
    { 
        for(int i = 0;i < n;i++) 
            scanf("%lf%lf",&p[i].x,&p[i].y); 
        sort(p,p+n,cmpxy); 
        printf("%.2lf
",Closest_Pair(0,n-1)/2); 
    } 
    return 0; 
} 

题解：LightOJ 1313 - Protect the Mines（凸包）