CF995C Leaving the Bar

题目描述

For a vector $v=(x,y)$ , define $|v|=\sqrt{x^2+y^2}$ .

Allen had a bit too much to drink at the bar, which is at the origin. There are $n$ vectors $v^1,v^2,\cdots ,v^n$ . Allen will make $n$ moves. As Allen's sense of direction is impaired, during the $i$ -th move he will either move in the direction $v^i$ or $-v^i$ . In other words, if his position is currently $p=(x,y)$ , he will either move to $p+v^i$ or $p-v^i$ .

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 $p$ satisfies $|p|\le 1.5\cdot 10^6$ so that he can stay safe.

输入输出格式

输入格式：

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

Each of the following lines contains two space-separated integers $x^i$ and $y^i$ , meaning that $v^i=(x^i,y^i)$ . We have that $|v^i|\le 10^6$ for all $i$ .

输出格式：

Output a single line containing $n$ integers $c^1,c^2,\cdots ,c^n$ , each of which is either $1$ or $-1$ . Your solution is correct if the value of $p={\sum }^{i=1}{c}^i{v}^i$ , satisfies $|p|\le 1.5\cdot 10^6$ .

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=100005;
const ll T=1500000*1ll*1500000;
ll ans[N];
ll n;
struct node{
    ll id,x,y;
}a[N];

il int gi(){
    int a=0;char x=getchar();bool f=0;
    while((x<'0'||x>'9')&&x!='-')x=getchar();
    if(x=='-')x=getchar(),f=1;
    while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    return f?-a:a;
}

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

int main(){
    srand(time(0));
    n=gi();
    For(i,1,n) a[i].x=gi(),a[i].y=gi(),a[i].id=i;
    ll x,y;
    while(1){
        random_shuffle(a+1,a+n+1);
        x=0,y=0;
        For(i,1,n) 
            if(lala(x-a[i].x,y-a[i].y)<lala(a[i].x+x,a[i].y+y)) ans[a[i].id]=-1,x-=a[i].x,y-=a[i].y;
            else ans[a[i].id]=1,x+=a[i].x,y+=a[i].y;
        if(lala(x,y)<=T) {For(i,1,n) printf("%lld ",ans[i]);break;}
    }
    return 0;
}