hdu_1392_Surround the Trees(凸包)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1392

题意：求凸包，不知道的百度

题解：模版题

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
/*  
* 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 = 110;
const double eps = 1e-8; 
const double PI = acos(-1.0); 
int sgn(double x) { 
 if(fabs(x) < eps)return 0; 
 if(x < 0)return -1;  
 else return 1; } 
struct Point {
     double x,y; 
     Point(){} 
     Point(double _x,double _y){x = _x,y = _y;} 
     Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}//叉积 
     double operator ^(const Point &b)const{return x*b.y-y*b.x;}//点积 
     double operator *(const Point &b)const{return x*b.x + y*b.y;}//绕原点旋转角度B（弧度值），后x,y的变化 
     void transXY(double B){double tx = x,ty = y,x = tx*cos(B) - ty*sin(B),y = tx*sin(B) + ty*cos(B);}
}list[MAXN];
int S[MAXN],top;//相对于list[0]的极角排序

double dist(Point a,Point b){return sqrt((a-b)*(a-b));}
bool _cmp(Point p1,Point p2){
    double tmp =(p1-list[0])^(p2-list[0]); 
    if(sgn(tmp)>0)return 1; 
    else if(sgn(tmp)==0&&sgn(dist(p1,list[0])-dist(p2,list[0]))<= 0)return 1;  
    return 0;
}
void Graham(int n){
    Point p0=list[0],tp; 
    int k=0;
    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;
    tp=list[k],list[k]=list[0],list[0]=tp,sort(list+1,list+n,_cmp);
    if(n==1){top=1,S[0]=0;return;} 
    if(n==2){top=2,S[0]=0,S[1]=1;return;} 
    S[0]=0,S[1]=1,top=2; 
    for(int i=2;i<n;i++){
        while(top>1&&sgn((list[S[top-1]]-list[S[top-2]])^(list[i]-list[S[top-2]]))<=0)top--;
        S[top++]=i; 
    } 
}

int main(){
    int n;
    while(~scanf("%d",&n),n){
        for(int i=0;i<n;i++)scanf("%lf%lf",&list[i].x,&list[i].y);
        if(n==1){puts("0.00");continue;}
        else if(n==2){printf("%.2lf
",dist(list[0],list[1]));continue;}
        Graham(n);
        double ans=0;
        for(int i=0;i<top-1;i++)ans+=dist(list[S[i]],list[S[i+1]]);
        ans+=dist(list[S[0]],list[S[top-1]]);
        printf("%.2lf
",ans);
    }
    return 0;
}