BZOJ 4080 Wf2014 Sensor Network randomization, bzojwf2014

BZOJ 4080 Wf2014 Sensor Network randomization, bzojwf2014

On a given plane N Point, find the largest point set, so that the distance between the two is not more D

It's useless to add any pruning to the search T ......

Good randomization Method

A random sequence is added in order, and then the answer is updated.

Is it said to be reliable? It's almost done after writing it.

#include <bitset>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 110using namespace std;typedef bitset<M> abcd;struct Point{    int x,y;    friend istream& operator >> (istream &_,Point &p)    {        return scanf("%d%d",&p.x,&p.y),_;    }    friend bool operator < (const Point &p1,const Point &p2)    {        return p1.x < p2.x ;    }    friend int Distance(const Point &p1,const Point &p2)    {        return (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) ;    }}points[M];int n,d;abcd a[M];abcd now,able,ans;int main(){    int i,j;    cin>>n>>d;    for(i=1;i<=n;i++)        cin>>points[i];    for(i=1;i<=n;i++)        for(j=i+1;j<=n;j++)            if(Distance(points[i],points[j])<=d*d)                a[i][j]=a[j][i]=true;    static int order[M];    for(i=1;i<=n;i++)    {        able[i]=true;        order[i]=i;    }    for(j=1;j<=1000;j++)    {        abcd able=::able;        now.reset();        for(i=1;i<=n;i++)            if(able[order[i]])            {                now[order[i]]=true;                able&=a[order[i]];            }        if(now.count()>ans.count())            ans=now;        random_shuffle(order+1,order+n+1);    }    cout<<ans.count()<<endl;    for(i=1;i<=n;i++)        if(ans[i])            printf("%d ",i);    return 0;}