POJ-1696-Space Ant（凸包变形）

题目大意：给出一些点，让你只能向左转向走完所有的点，输出路径。
思路：按着要求模拟即可，走过的点标记一下，然后跑凸包即可。
ACCode：
//#pragma comment(linker, "/STACK:1024000000,1024000000")
  
#include<stdio.h>
#include<string.h> 
#include<math.h> 
   
#include<map>  
#include<set>
#include<deque> 
#include<queue> 
#include<stack> 
#include<bitset>
#include<string> 
#include<fstream>
#include<iostream> 
#include<algorithm> 
using namespace std; 
  
#define ll long long 
#define Pair pair<int,int>
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// ??
//std::ios::sync_with_stdio(false);
//  register
const int MAXN=1e2+10;
const int INF32=0x3f3f3f3f;
const ll INF64=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);
const double EPS=1.0e-8;

struct Point{
	double x,y,t,d;int id;
	Point(double _x=0,double _y=0,int _id=0,double _t=0,double _d=0){
		x=_x;y=_y;t=_t;d=_d;id=_id;
	}
	friend Point operator + (const Point &a,const Point &b){
		return Point(a.x+b.x,a.y+b.y);
	}
	friend Point operator - (const Point &a,const Point &b){
		return Point(a.x-b.x,a.y-b.y);
	}
	friend double operator ^ (Point a,Point b){//??????? 
		return a.x*b.y-a.y*b.x;
	}
	friend int operator == (const Point &a,const Point &b){
		if(fabs(a.x-b.x)<EPS&&fabs(a.y-b.y)<EPS) return 1;
		return 0;
	}
};
struct V{
	Point start,end;double ang;
	V(Point _start=Point(0,0),Point _end=Point(0,0),double _ang=0.0){
		start=_start;end=_end;ang=_ang;
	}
	friend V operator + (const V &a,const V &b){
		return V(a.start+b.start,a.end+b.end);
	}
	friend V operator - (const V &a,const V &b){
		return V(a.start-b.start,a.end-b.end);
	}
};
Point Dots[MAXN];
Point Stk[MAXN];int Top;
int Vis[MAXN];
int n;

int LineInter(V l1,V l2){//--判断点相交 
	if(max(l1.start.x,l1.end.x)>=min(l2.start.x,l2.end.x)&&
	max(l2.start.x,l2.end.x)>=min(l1.start.x,l1.end.x)&&
	max(l1.start.y,l1.end.y)>=min(l2.start.y,l2.end.y)&&
	max(l2.start.y,l2.end.y)>=min(l1.start.y,l1.end.y)){
		if(((l2.end-l2.start)^(l1.start-l2.start))*((l2.end-l2.start)^(l1.end-l2.start))<=0&&
		((l1.end-l1.start)^(l2.start-l1.start))*((l1.end-l1.start)^(l2.end-l1.start))<=0)
			//printf(" 相交： l1:(%lf,%lf),(%lf,%lf) l2:(%lf,%lf),(%lf,%lf)\n",l1.start.x,l1.start.y,l1.end.x,l1.end.y,l2.start.x,l2.start.y,l2.end.x,l2.end.y);
			return 1;
	}//printf("不相交： l1:(%lf,%lf),(%lf,%lf) l2:(%lf,%lf),(%lf,%lf)\n",l1.start.x,l1.start.y,l1.end.x,l1.end.y,l2.start.x,l2.start.y,l2.end.x,l2.end.y);
	return 0;
}
int Judge(V l){
	for(int i=2;i<Top;++i){
		if(LineInter(l,V(Stk[i],Stk[i-1]))) return 1;
	}return 0;
}
int Cmp(Point a,Point b){
	if(fabs(a.y-b.y)<EPS){
		return a.x<b.x;
	}return a.y<b.y;
}
int main(){
	int T;scanf("%d",&T);
	while(T--){
		clean(Vis,0);
		scanf("%d",&n);
		double x,y;int id;
		for(int i=1;i<=n;++i){
			scanf("%d%lf%lf",&id,&x,&y);
			Dots[i]=Point(x,y,id);
		}sort(Dots+1,Dots+1+n,Cmp);
		Top=0;
		Stk[++Top]=Point(0,Dots[1].y);Stk[++Top]=Dots[1];
		Vis[Stk[Top].id]=1;
		for(int i=1;i<=n;++i){
			for(int j=1;j<=n;++j){
				if(Vis[Dots[j].id]) continue;
				while(Top>=2&&(((Dots[j]-Stk[Top])^(Stk[Top]-Stk[Top-1]))>EPS||Judge(V(Stk[Top],Dots[j])))){
					Vis[Stk[Top].id]=0;--Top;
				}
				Stk[++Top]=Dots[j];Vis[Stk[Top].id]=1;
//				cout<<"add : "<<char(Dots[j].id+'A'-1)<<endl;
//				for(int k=2;k<=Top;++k) cout<<char(Stk[k].id+'A'-1)<<" ";cout<<endl;
			}
//			cout<<"DeBug:-----"<<endl;
//			for(int j=2;j<=Top;++j) cout<<char(Stk[j].id+'A'-1)<<" ";cout<<endl;
		}printf("%d %d",Top-1,Stk[2].id);
		for(int i=3;i<=Top;++i){
			printf(" %d",Stk[i].id);
		}printf("\n");
	}
	
}

/*
1
14
1 6 11
2 11 9
3 8 7
4 12 8
5 9 20
6 3 2
7 1 6
8 2 13
9 15 1
10 14 17
11 13 19
12 5 18
13 7 3
14 10 16

*/
POJ-1696-Space Ant（凸包变形）

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