POJ 1696

题目链接：http://poj.org/problem?id=1696

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题目思路：

简单凸包，用的叉积擂台赛，注意当有共线点时的处理。一遍ac

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源代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <math.h>
#include <queue>
#include <algorithm>

using namespace std;
#define INF 10000000000
#define min(a,b) ((a)>(b)?(b):(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define maxn 110
const double eps = 1e-10;

typedef double T;

int n = 0;

struct Pt
{
   T x;
   T y;
   int index;
   int flag;

   Pt(){}
   Pt(T px,T py) { x = px; y = py; }

   Pt operator +(const Pt &p) { return Pt(x+p.x,y+p.y); }

   Pt operator -(const Pt &p) { return Pt(x-p.x,y-p.y); }

   int operator ==(const Pt &p)
   {
       T temp = 0;
       temp = sqrt((x - p.x)*(x - p.x)+(y-p.y)*(y-p.y));

       if(temp<eps)
         return 1;
       else
         return 0;
   }
};

struct Line
{
    Pt a;
    Pt b;
    int num;

    Line(){}
    Line (Pt p1,Pt p2) { a = p1;  b = p2;}
};

//ab * cd
T dpr(Pt a,Pt b,Pt c,Pt d)  { return (b.x-a.x)*(d.x-c.x)+(b.y-a.y)*(d.y-c.y);}
T dpr_three(Pt a,Pt b,Pt c) { return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);}

//ab × cd
T cpr(Pt a,Pt b,Pt c,Pt d){ return (b.x-a.x)*(d.y-c.y)-(b.y-a.y)*(d.x-c.x);}
T cpr_three(Pt o,Pt b,Pt c) { return (b.x-o.x)*(c.y-o.y)-(b.y-o.y)*(c.x-o.x);}

//a b为向量，向量的点积和叉积
T det(const Pt& a,const Pt& b) { return a.x*b.y - a.y*b.x; }
T dot(const Pt& a,const Pt& b) { return a.x*b.x + a.y*b.y; }

T d(Pt a,Pt b) { return sqrt((a.x - b.x)*(a.x - b.x) + (a.y-b.y)*(a.y - b.y));}

//-------------符号函数--------------------
int sgn(T x) { return x<-eps?-1:x>eps;}

int epsfun(T x)
{
    if(x<-eps || x<eps)
      return 0;
    else
      return x;
}

//bool cmp(Line l1,Line l2) { return l1.a.x < l2.a.x; }
bool cmp(Pt p1,Pt p2)
{
     if(p1.y == p2.y) return p1.x<p2.x;
     return p1.y<p2.y;
}

void judge_line(Line line1 ,Line line2)    //判断两直线位置关系并求交点
{
     if(sgn(cpr(line1.b,line1.a,line2.b,line2.a) == 0))
     {
        if(sgn(cpr(line1.b,line1.a,line2.b,line1.a) == 0))
          printf("LINE\n");
        else
          printf("NONE\n");
     }
     else
     {
        T s1 = cpr(line1.b,line1.a,line2.a,line1.a),s2 = cpr(line1.b,line1.a,line2.b,line1.a);
        printf("POINT ");
        printf("%.2f %.2f\n",(line2.a.x*s2-line2.b.x*s1)/(s2-s1),(line2.a.y*s2-line2.b.y*s1)/(s2-s1));
     }
}
//-------------直线线段相交--------------------
//直线与线段 规范相交： Line a为直线，Line b为线段
int line_seg_cross(Line a,Line b)
{
    if(sgn(cpr(a.b,a.a,b.a,a.a))*sgn(cpr(a.b,a.a,b.b,a.a))<0)
      return 1;
    else
      return 0;
}

int res_line_seg_cross()
{

}

//-------------线段线段相交--------------------
int between(Pt a,Pt b,Pt c)
{
    if(dpr_three(a,c,b)<=0)
      return 1;
    return 0;
}

//非规范相交（包括多个交点）
int segcross(Line linea,Line lineb)
{
    T s1 = 0,s2 = 0,s3 = 0,s4 = 0;
    int d1 = 0,d2 = 0,d3 = 0,d4 = 0 ;
    Pt a = linea.a,b = linea.b,c = lineb.a,d = lineb.b;

    d1 = sgn(s1 = cpr_three(a,b,c));
    d2 = sgn(s2 = cpr_three(a,b,d));
    d3 = sgn(s3 = cpr_three(c,d,a));
    d4 = sgn(s4 = cpr_three(c,d,b));

    if((d1^d2) == -2 && (d3^d4) == -2)  return 1;

    if((d1 == 0 && between(c,a,b)) || (d2 == 0 && between(d,a,b)) || (d3 == 0 && between(a,c,d)) || (d4 == 0 && between(b,c,d)))
        return 1;

    return 0;
}

//判规范相交
bool res_segcross(Line a,Line b)
{
    int d1 = sgn(det(a.a-b.a,b.b-b.a)),
        d2 = sgn(det(a.b-b.a,b.b-b.a)),
        d3 = sgn(det(b.a-a.a,a.b-a.a)),
        d4 = sgn(det(b.b-a.a,a.b-a.a));

    if (d1*d2 == -1 && d3*d4 == -1)
        return true;

    return false;
}

//规范相交：1， 非规范相交且交点有无数个：3 非规范相交且交点唯一：2  不相交：0

int segcross_all(Pt a,Pt b,Pt c,Pt d,Pt *p)   //规范相交：1， 非规范相交且交点有无数个：3 非规范相交且交点唯一：2  不相交：0
{
    T s1 = 0,s2 = 0,s3 = 0,s4 = 0;
    int d1 = 0,d2 = 0,d3 = 0,d4 = 0 ;

    d1 = sgn(s1 = cpr_three(a,b,c));
    d2 = sgn(s2 = cpr_three(a,b,d));
    d3 = sgn(s3 = cpr_three(c,d,a));
    d4 = sgn(s4 = cpr_three(c,d,b));

    if((d1^d2) == -2 && (d3^d4) == -2)
    {
        (*p).x = (c.x*s2 - d.x*s1)/(s2 - s1);
        (*p).y = (c.y*s2 - d.y*s1)/(s2 - s1);
        return 1;
    }
    if(!(d1 || d2 || d3 ||d4) && (between(c,a,b)||between(d,a,b)||between(a,c,d)||between(b,c,d)))
       return 3;

    if(d1 == 0 && between(c,a,b))
    {
        (*p).x = c.x;
        (*p).y = c.y;
        return 2;
    }
    if(d2 == 0 && between(d,a,b))
    {
        (*p).x = d.x;
        (*p).y = d.y;
        return 2;
    }
    if(d3 == 0 && between(a,c,d))
    {
        (*p).x = a.x;
        (*p).y = a.y;
        return 2;
    }
    if(d4 == 0 && between(b,c,d))
    {
        (*p).x = b.x;
        (*p).y = b.y;
        return 2;
    }

    return 0;
}

Pt point[maxn];

int main()
{
   // freopen("in.in", "r", stdin);
   // freopen("out.txt","w",stdout);
    int m = 0,n = 0;
    int i = 0,j = 0;
    int x = 0,y = 0,cur = 0;

    scanf("%d",&m);
    while(m--)
    {
        scanf("%d",&n);
        for(i = 0;i<n;i++)
        {
            scanf("%d%d%d",&cur,&x,&y);
            point[i] = Pt(x,y);
            point[i].index = cur;
            point[i].flag = 0;
        }
        sort(point,point+n,cmp);
        point[0].flag = 1;
        printf("%d %d ",n,point[0].index);
        cur  = 0;

        for(i = 1;i<n;i++)
        {
            int bst = -1;
            for(j = 1;j<n;j++)
            {
                if(point[j].flag)
                   continue;
                if(bst == -1)
                  bst = j;
                else
                {
                    if(cpr_three(point[cur],point[bst],point[j]) == 0)
                    {
                        if(point[bst].x>point[j].x)
                          bst = j;
                    }
                    else if(cpr_three(point[cur],point[bst],point[j])<0)
                           bst = j;
                }
            }
           if(i == n-1)
              printf("%d\n",point[bst].index);
           else
              printf("%d ",point[bst].index);
           cur = bst;
           point[bst].flag = 1;
        }
    }

    return 0;
}