[Codeforces995C]Leaving the Bar 瞎搞

大致题意：

　　给出平面上n个向量，对于每个向量可以选择正的V或负的-V，求按照选择的向量走完，最后距离原点<=1.5*1e6的一个选择方案

　　非正解！！！！！！！！！！

　　先按距离原点距离由远到近贪心，贪完后答案若不满足，则进行一次dfs找出正确地路线即可（数据较水）

　　也可以每次每次讲向量随机排序然后贪心，直到贪到正确答案为止。

　　

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 100100
#define eps 1e-9
#define For(i,a,b) for(int i=a;i<=b;i++)
#define Fore(i,a,b) for(int i=a;i>=b;i--)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fre freopen
#define pi acos(-1.0)
#define inf 1e6+7
#define Vector Point
const int Mod=1e9+;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return ;
    return x<?-:;
}
struct Point{
    double x,y;
    int id;
    Point(double x=,double y=):x(x),y(y) {}
    bool operator < (const Point &a)const{
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
    Point operator - (const Point &a)const{
        return Point(x-a.x,y-a.y);
    }
    Point operator + (const Point &a)const{
        return Point(x+a.x,y+a.y);
    }
    Point operator * (const double &a)const{
        return Point(x*a,y*a);
    }
    Point operator / (const double &a)const{
        return Point(x/a,y/a);
    }
    void read(){
        scanf("%lf%lf",&x,&y);
    }
    void out(){
        cout<<"debug: "<<x<<" "<<y<<endl;
    }
    bool operator == (const Point &a)const{
        return dcmp(x-a.x)== && dcmp(y-a.y)==;
    }
};
double Dot(Vector a,Vector b) {
    return a.x*b.x+a.y*b.y;
}
double dis(Vector a) {
    return sqrt(Dot(a,a));
}
double Cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
int Maxx=;
bool cmp(Point a,Point b){
    return dcmp(dis(a)-dis(b))>;
}
bool check(Point o,Point p1){
    return dcmp(dis(o+p1)-dis(o-p1))<=;
}
int ans[];
Point p[];
int n,mk;
void dfs(int now,Point o){
    if(mk) return ;
    if(now==n) {
        if(dcmp(dis(o)-Maxx)>) return ;
        mk=;
        For(i,,n-) {
            if(i) printf(" ");
            printf("%d",ans[i]);
        }
        cout<<endl;
        return ;
    }
    ans[p[now].id]=;
    dfs(now+,o+p[now]);
    ans[p[now].id]=-;
    dfs(now+,o-p[now]);
}
void solve(){
    cin>>n;
    For(i,,n-) p[i].read(),p[i].id=i;
    mk=;
    sort(p,p+n,cmp);
    Point o(,);
    For(i,,n-){
        if(check(o,p[i])) ans[p[i].id]=,o=o+p[i];
        else ans[p[i].id]=-,o=o-p[i];
    }
    if(dcmp(dis(o)-Maxx)<=) {
        For(i,,n-){
            if(i) printf(" ");
            printf("%d",ans[i]);
        }
        cout<<endl;
        return ;
    }
    dfs(,Point(,));
}
int main(){
//    fre("in.txt","r",stdin);
    int t=;
     solve();
    return ;
}

贪心+dfs