CodeForcesdiv1：995C - Leaving the Bar（随机算法+贪心）

For a vector $\stackrel{}{}$

Allen had a bit too much to drink at the bar, which is at the origin. There are $\mathrm{nn vectors \to v1,\to v2,\cdots ,\to vnv1\to ,v2\to ,\cdots ,vn\to .\; Allen\; will\; make nn moves.\; As\; Allen\text{'}s\; sense\; of\; direction\; is\; impaired,\; during\; the ii-th\; move\; he\; will\; either\; move\; in\; the\; direction \to vivi\to or -\to vi-vi\to .\; In\; other\; words,\; if\; his\; position\; is\; currently p=(x,y)p=(x,y),\; he\; will\; either\; move\; to p+\to vip+vi\to or p-\to vip-vi\to .}$

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 $\mathrm{pp satisfies |p|\le 1.5\cdot 106|p|\le 1.5\cdot 106 so\; that\; he\; can\; stay\; safe.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; moves.}$

Each of the following lines contains two space-separated integers ${\mathrm{xixi and yiyi,\; meaning\; that \to vi=(xi,yi)vi\to =(xi,yi).\; We\; have\; that |vi|\le 106|vi|\le 106 for\; all ii.}}^{}$

Output

Output a single line containing $\mathrm{nn integers c1,c2,\cdots ,cnc1,c2,\cdots ,cn,\; each\; of\; which\; is\; either 11 or -1-1.\; Your\; solution\; is\; correct\; if\; the\; value\; of p=\sum ni=1ci\to vip=\sum i=1ncivi\to ,\; satisfies |p|\le 1.5\cdot 106|p|\le 1.5\cdot 106.}$

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

Examples

input

Copy

3
999999 0
0 999999
999999 0

output

Copy

1 1 -1 

input

Copy

1
-824590 246031

output

Copy

1 

input

Copy

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

output

Copy

1 1 1 1 1 1 1 -1 

题意：给定一些向量，你可以改变它的符号，使得这些向量之和的长度小于1.5e6。

思路：考虑到每条边的长度小于1e6，所以任意两条边的向量和长度小于1.414e6<1.5e6。任意三条边的向量和（可以改变方向）的长度也成立，下面给出证明...

所以我们贪心, 保证向量和离原点更近. 然后下面的代码AC了,然后又被FST了.是因为每次我贪心原则是一样的.最后的结果有可能大于1.5e6.  我们需要加一些随机性, 多次贪心,直到结果满足题意.（这里随机的作用就算多次贪心，直到满足）

证明：每三个向量abc中都能找到两个向量合起来 <= 1e6，然后不断合并，最后只会剩下一个或者两个向量。

对于a和b，1，若ab小于60度，其中一个转向，则变为大于120度，<1e6；2，ab角度大于120度，则其向量和的长度小于1e6。3，角度在60到120度之间，那么第三边c，不论怎么放，都或与a或与b满足上面的关系，使得其向量和长度小于1e6，那么一直这样合并，最后只剩下一边或者两边，而两边的情况最坏为1.414e6

FST代码,WA79:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1000000;
const ll p=1500000;
ll ans[maxn+10],x[maxn+10],y[maxn+10];
ll X,Y,P;
int main()
{
    int N,i,j; P=p*p;scanf("%d",&N);
    for(i=1;i<=N;i++){
       scanf("%I64d%I64d",&x[i],&y[i]);
       if(X>=0&&Y>=0&&x[i]<=0&&y[i]<=0) X+=x[i],Y+=y[i],ans[i]=1;
       else if(X>=0&&Y>=0&&x[i]>=0&&y[i]>=0) X-=x[i],Y-=y[i],ans[i]=-1;
       else if(X<=0&&Y<=0&&x[i]<=0&&y[i]<=0) X-=x[i],Y-=y[i],ans[i]=-1;
       else if(X<=0&&Y<=0&&x[i]>=0&&y[i]>=0) X+=x[i],Y+=y[i],ans[i]=1;
       else if((X+x[i])*(X+x[i])+(Y+y[i])*(Y+y[i])<(X-x[i])*(X-x[i])+(Y-y[i])*(Y-y[i])) X+=x[i],Y+=y[i],ans[i]=1;
       else X-=x[i],Y-=y[i],ans[i]=-1;
    }
    for(i=1;i<=N;i++) printf("%d ",ans[i]);
    return 0;
}

AC代码:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1000000;
const ll p=1500000;
ll X,Y,P; int ans[maxn];
struct in{ ll x,y,id; }s[maxn];
int main()
{
    int N,i,j; P=p*p;scanf("%d",&N);
    for(i=1;i<=N;i++){
       scanf("%I64d%I64d",&s[i].x,&s[i].y); s[i].id=i;
    }
    while(true){
       random_shuffle(s+1,s+N+1); X=Y=0;
       for(i=1;i<=N;i++){        
          if((X+s[i].x)*(X+s[i].x)+(Y+s[i].y)*(Y+s[i].y)>(X-s[i].x)*(X-s[i].x)+(Y-s[i].y)*(Y-s[i].y)) X-=s[i].x,Y-=s[i].y,ans[s[i].id]=-1;
          else X+=s[i].x,Y+=s[i].y,ans[s[i].id]=1;
       }
       if(X*X+Y*Y<=P) {
               for(i=1;i<=N;i++) printf("%d ",ans[i]); return 0;
       }
    }
    return 0;
}

 

 

 （div1E也是需要随机算法！！！