CF995C Leaving the Bar

题目描述

For a vector v⃗=(x,y) \vec{v} = (x, y) v=(x,y) , define ∣v∣=x2+y2 |v| = \sqrt{x^2 + y^2} ∣v∣=x2+y2​ .

Allen had a bit too much to drink at the bar, which is at the origin. There are n n n vectors v1⃗,v2⃗,⋯,vn⃗ \vec{v_1}, \vec{v_2}, \cdots, \vec{v_n} v1​​,v2​​,⋯,vn​​ . Allen will make n n n moves. As Allen's sense of direction is impaired, during the i i i -th move he will either move in the direction vi⃗ \vec{v_i} vi​​ or −vi⃗ -\vec{v_i} −vi​​ . In other words, if his position is currently p=(x,y) p = (x, y) p=(x,y) , he will either move to p+vi⃗ p + \vec{v_i} p+vi​​ or p−vi⃗ p - \vec{v_i} p−vi​​ .

Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p p p satisfies ∣p∣≤1.5⋅106 |p| \le 1.5 \cdot 10^6 ∣p∣≤1.5⋅106 so that he can stay safe.

输入输出格式

输入格式：

The first line contains a single integer n n n ( 1≤n≤105 1 \le n \le 10^5 1≤n≤105 ) — the number of moves.

Each of the following lines contains two space-separated integers xi x_i xi​ and yi y_i yi​ , meaning that vi⃗=(xi,yi) \vec{v_i} = (x_i, y_i) vi​​=(xi​,yi​) . We have that ∣vi∣≤106 |v_i| \le 10^6 ∣vi​∣≤106 for all i i i .

输出格式：

Output a single line containing n n n integers c1,c2,⋯,cn c_1, c_2, \cdots, c_n c1​,c2​,⋯,cn​ , each of which is either 1 1 1 or −1 -1 −1 . Your solution is correct if the value of p=∑i=1ncivi⃗ p = \sum_{i = 1}^n c_i \vec{v_i} p=∑i=1n​ci​vi​​ , satisfies ∣p∣≤1.5⋅106 |p| \le 1.5 \cdot 10^6 ∣p∣≤1.5⋅106 .

It can be shown that a solution always exists under the given constraints.

输入输出样例

输入样例#1：

3
999999 0
0 999999
999999 0

输出样例#1：

1 1 -1

输入样例#2：

1
-824590 246031

输出样例#2：

1

输入样例#3：

8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899

输出样例#3：

1 1 1 1 1 1 1 -1

Solution:

　　本题很玄学，正解不会，直接随机。

　　用random_shuffle去随机打乱数组，然后贪心，对于第$i$个向量直接在$+1,-1$中选一个使向量长度小的，然后判断向量和的长度是否满足条件就好了。

代码：

 #include<bits/stdc++.h>
 #define il inline
 #define ll long long
 #define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
 #define Bor(i,a,b) for(int (i)=(b);(i)>=(a);(i)--)
 using namespace std;
 const int N=;
 const ll T=*1ll*;
 ll ans[N];
 ll n;
 struct node{
     ll id,x,y;
 }a[N];

 il int gi(){
     int a=;char x=getchar();bool f=;
     while((x<''||x>'')&&x!='-')x=getchar();
     if(x=='-')x=getchar(),f=;
     while(x>=''&&x<='')a=(a<<)+(a<<)+x-,x=getchar();
     return f?-a:a;
 }

 il ll lala(ll x,ll y){return x*x+y*y;}

 int main(){
     srand(time());
     n=gi();
     For(i,,n) a[i].x=gi(),a[i].y=gi(),a[i].id=i;
     ll x,y;
     while(){
         random_shuffle(a+,a+n+);
         x=,y=;
         For(i,,n)
             if(lala(x-a[i].x,y-a[i].y)<lala(a[i].x+x,a[i].y+y)) ans[a[i].id]=-,x-=a[i].x,y-=a[i].y;
             else ans[a[i].id]=,x+=a[i].x,y+=a[i].y;
         if(lala(x,y)<=T) {For(i,,n) printf("%lld ",ans[i]);break;}
     }
     return ;
 }