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 110const 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 * cdT 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 × cdT 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：0int 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;}