POJ 3714 平面最近点对

题意：

分为两种点，求两种点之间的平面最近点对

题解：

分治法。这个怎么暴力分治都能过。。

吐槽：我用了g++ tle+wa，毛线。。。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>

#define N 222222
#define EPS 1e-7
#define INF 1e15

using namespace std;

struct PO
{
    bool fg;
    double x,y;
}p[N];

int n,q1[N],q2[N];

inline bool cmp(const PO &a,const PO &b)
{
    return a.x<b.x;
}

inline int doublecmp(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

inline double get_dis(PO &a,PO &b)
{
    if(a.fg==b.fg) return INF;
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline double min(double a,double b)
{
    if(doublecmp(a-b)<=0) return a;
    return b;
}

inline void read()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) 
    {
        scanf("%lf%lf",&p[i].x,&p[i].y);
        p[i].fg=1;
    }
    for(int i=n+1;i<=(n<<1);i++)
    {
        scanf("%lf%lf",&p[i].x,&p[i].y);
        p[i].fg=0;
    }
    n<<=1;
}

inline double getmindis(int l,int r)
{
    if(l==r) return INF;
    if(l+1==r) return get_dis(p[l],p[r]);
    int mid=(l+r)>>1;
    double lmindis=getmindis(l,mid);
    double rmindis=getmindis(mid+1,r);
    double mindis=min(lmindis,rmindis);
    int cnt1=0,cnt2=0;
    
    for(int i=mid;i>=l;i--)
    {
        if(doublecmp(mindis-(p[mid].x-p[i].x))<=0) break;
        else if(doublecmp(mindis-fabs(p[mid].y-p[i].y))>=0) q1[++cnt1]=i;
    }
    for(int i=mid+1;i<=r;i++)
    {
        if(doublecmp(mindis-(p[i].x-p[mid].x))<=0) break;
        else if(doublecmp(mindis-fabs(p[mid].y-p[i].y))>=0) q2[++cnt2]=i;
    }
    for(int i=1;i<=cnt1;i++)
        for(int j=1;j<=cnt2;j++)
            mindis=min(mindis,get_dis(p[q1[i]],p[q2[j]]));
    return mindis;
}

inline void go()
{
    sort(p+1,p+1+n,cmp);
    printf("%.3lf\n",getmindis(1,n));
}

int main()
{
    int cas; scanf("%d",&cas);
    while(cas--) read(),go();
    return 0;
}

 上面那个跑的慢

这题考虑精度反而跪了。。。。无语。。。

ZOJ 2107/HDU 1007

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>

#define N 2222222
#define EPS 1e-7
#define INF 1e10

using namespace std;

struct PO
{
    double x,y;
}p[N],q[N];

int n;

inline bool cmpx(const PO &a,const PO &b)
{
    return a.x<b.x;
}

inline bool cmpy(const PO &a,const PO &b)
{
    return a.y<b.y;
}

inline int doublecmp(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

inline double get_dis(PO &a,PO &b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline double min(double a,double b)
{
    if(doublecmp(a-b)<=0) return a;
    return b;
}

inline void read()
{
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
}

inline double getmindis(int l,int r)
{
    if(l+1==r) return get_dis(p[l],p[r]);
    if(l+2==r) return min(get_dis(p[l],p[l+1]),min(get_dis(p[l+1],p[r]),get_dis(p[l],p[r])));
    int mid=(l+r)>>1;
    double mindis=min(getmindis(l,mid),getmindis(mid+1,r));
    int cnt=0;
    for(int i=l;i<=r;i++)
        if(doublecmp(mindis-fabs(p[i].x-p[mid].x))>=0) q[++cnt]=p[i];
    sort(q+1,q+1+cnt,cmpy);
    for(int i=1;i<=cnt;i++)
        for(int j=i+1;j<=cnt;j++)
        {
            if(q[j].y-q[i].y>=mindis) break;
            //if(doublecmp(p[j].y-p[i].y-mindis)>=0) break;
            mindis=min(mindis,get_dis(q[i],q[j]));
        }
    return mindis;
}

inline void go()
{
    sort(p+1,p+1+n,cmpx);
    printf("%.2lf\n",getmindis(1,n)*0.5);
}

int main()
{
    while(scanf("%d",&n),n) read(),go();
    return 0;
}