HDU6325 (Interstellar Travel)[凸包，字典序]

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6325

注意观察每次移动的花费xi yj - xj yi，这是一个叉积。题目希望花费的负值最大，也就是所有的叉积和尽可能小。贪心考虑，我们需要让移动路线尽可能「上凸」，这样可以向顺时针转动的角度也会尽可能的大。于是维护一个上凸包即可。注意字典序的处理方式，拓展凸包时，如果在一个方向上遇到字典序较大的点，则不要弹出栈里的当前点。

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const double eps=1e-8;
const double inf=1e20;
const double pi=acos(-1.0);
const int maxn=200009;
int sgn(double x) {
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    else return 1;
}
inline double sqr(double x) { return x*x; }
struct Point {
    double x,y;
    int index;
    Point() {}
    Point(double _x,double _y,int _index=0) {
        x=_x;
        y=_y;
        index=_index;
    }
    double distance(Point p) {
        return hypot(x-p.x,y-p.y);
    }
    double len() {
        return hypot(x,y);
    }
    Point trunc(double r) {
        double l=len();
        if(!sgn(l)) return *this;
        r/=l;
        return Point(x*r,y*r);
    }
    bool operator ==(Point b) const {
        return sgn(x-b.x)==0 && sgn(y-b.y)==0;
    }
//    bool operator <(Point b) const {
//        return sgn(x-b.x)==0 ? sgn(y-b.y)<0:x<b.x;
//    }
    Point operator -(const Point& b) const {
        return Point(x-b.x,y-b.y);
    }
    Point operator +(const Point& b) const {
        return Point(x+b.x,y+b.y);
    }
    double operator ^(const Point& b) const {
        return x*b.y-y*b.x;
    }
    double operator *(const Point& b) const {
        return x*b.x+y*b.y;
    }
    Point rotleft() {
        return Point(-y,x);
    }
    Point rotright() {
        return Point(y,-x);
    }
};
struct polygon {
    int n;
    Point p[maxn];
    void add(Point q) {
        p[n++]=q;
    }
    void init() {
        n=0;
    }
    struct cmp {
        Point p;
        cmp(const Point& p0) { p=p0; }
        bool operator ()(const Point& aa,const Point& bb) {
            Point a=aa,b=bb;
            int d=sgn((a-p)^(b-p));
            if(d==0) {
                return sgn(a.distance(p)-b.distance(p)) < 0;
            }
            return d < 0;
        }
    };
    void norm() {
        Point mi=p[0];
        sort(p,p+n,cmp(mi));
    }
    void print() {
        for(int i=0;i<n;i++) {
            printf(i==n-1?"%d":"%d ",p[i].index);
        }
        printf("\n");
    }
    void Graham(polygon &convex) {
        norm();
        int &top=convex.n;
        top=0;
        if(n==1) {
            top=1;
            convex.p[0]=p[0];
            return ;
        }
        if(n==2) {
            top=2;
            convex.p[0]=p[0];
            convex.p[1]=p[1];
            if(convex.p[0]==convex.p[1]) top--;
            return ;
        }
        convex.p[0]=p[0];
        convex.p[1]=p[1];
        top=2;
        for(int i=2;i<n;i++) {
            while(top>1 && sgn((convex.p[top-1]-convex.p[top-2])
                ^(p[i]-convex.p[top-2]))>=0) {
                    if(sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2]))==0
                       && convex.p[top-1].index<p[i].index) break;
                    top--;
                }
            if(!(convex.p[top-1]==p[i]))
                convex.p[top++]=p[i];
        }
        if(convex.n==2 && (convex.p[0]==convex.p[1])) convex.n--;
    }
}poly,con;
int main() {
    int T;
    int n;
    int x,y;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        poly.init();
        for(int i=1;i<=n;i++) {
            scanf("%d%d",&x,&y);
            poly.add(Point(x,y,i));
        }
        con.init();
        poly.Graham(con);
        con.print();
    }
}