LightOj1285

传送门

题意：给你一些点，然后把它们用一条线把它们连起来，构成一个多边形，不能有相交，必须用完所有的点，如果不能构成输出Impossible；

不能构成就是所有的点在一条直线上的时候；先按极角进行排序，然后倒着找到一个不再起点到终点那条线上的点，倒着连接起来；

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=4e3+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;int dcmp(double x){    if(fabs(x) < EPS) return 0;    else return x < 0 ? -1 : 1;}struct Point{    double x, y;    int id;    //Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //¿½±´¹¹Ô캯Êý    Point(double x = 0, double y = 0) : x(x), y(y) { }    inline void input()    {        scanf("%lf%lf",&x,&y);    }    inline void print()    {        printf("%.6lf %.6lf\n",x,y);    }    bool operator == (const Point& e) const    {        return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0;    }    Point operator + (Point q){ return Point(x+q.x,y+q.y);}    Point operator - (Point q){ return Point(x-q.x,y-q.y);}    Point operator * (double q){ return Point(x*q,y*q);}    Point operator / (double q){ return Point(x/q,y/q);}    Point &operator +=(Point q){ x+=q.x;y+=q.y; return *this;}    Point &operator -=(Point q){ x-=q.x;y-=q.y; return *this;}    double operator *(const Point& q) const{        return x*q.x+y*q.y;    }    double operator ^(const Point& q) const{        return x*q.y-y*q.x;    }    double len()    {        return sqrt(x*x+y*y);    }    void read()    {        int xx,yy;        scanf("%d%d",&xx,&yy);        x=xx,y=yy;    }}Q[2010];int n;bool cmp(Point p1,Point p2)//位置排序，找到最下方的；{    if(p1.y!=p2.y)        return p1.y<p2.y;    return p1.x<p2.x;//若有多个下方的找左边的；}Point polor;bool polor_cmp(Point a,Point b){    double tmp=(a-polor)^(b-polor);    if(dcmp(tmp)==0)        return (a-polor).len()<(b-polor).len();    else if(dcmp(tmp)<0) return false;    else return true;}void solve(){    s_1(n);    FOr(0,n,i)    {        Q[i].read();        Q[i].id=i;    }    sort(Q,Q+n,cmp);    polor=Q[0];    sort(Q+1,Q+n,polor_cmp);    int tmp=0;    for(int i=n-2;i>0;i--)    {        if(((Q[n-1]-Q[0])^(Q[i]-Q[0]))!=0)        {            tmp=i;            break;        }    }    if(tmp==0)    {        puts("Impossible");        return ;    }    reverse(Q+tmp+1,Q+n);    FOr(0,n,i)    {        printf("%d%c",Q[i].id,i==n-1?'\n':' ');    }}int main(){   // freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    s_1(t);    for(int cas=1;cas<=t;cas++)    {        printf("Case %d:\n",cas);        solve();    }}