P4423 [BJWC2011] 最小三角形 与 SP7209 CLOSEST - Closest Triplet

noi 模拟赛 t1，所以打了些部分分，不介意吧……

思路：

仿照平面最近点对思路，先按照横坐标排序，考虑分治。

对于分割线 \(y=X\)，考虑求跨过这条线的贡献，设 \(d\) 为左边和右边分治结果的最小值，则这三点中最长边的长度必须 \(\le \frac{d}{2}\)，不然不会比 \(d\) 更优。

则我们只需要考虑横坐标到分割线的距离 \(\le \frac{d}{2}\) 的贡献，将这些点找出来，再按照纵坐标进排序，这里使用归并排序的话可以降一个 \(\log\)，但是没必要。

然后暴力枚举这 \(3\) 个点，使得三个点的 \(y\) 坐标的极差 \(\le \frac{d}{2}\)，然后计算贡献即可。

时间复杂度为 \(O(N \log^2 N)\)。

P4423 [BJWC2011] 最小三角形 Code：

#include<bits/stdc++.h>
#define Add(x,y) (x+y>=mod)?(x+y-mod):(x+y)
#define lowbit(x) x&(-x)
#define pi pair<ll,ll>
#define pii pair<ll,pair<ll,ll>>
#define iip pair<pair<ll,ll>,ll>
#define ppii pair<pair<ll,ll>,pair<ll,ll>>
#define fi first
#define se second
#define full(l,r,x) for(auto it=l;it!=r;it++) (*it)=x
#define Full(a) memset(a,0,sizeof(a))
#define open(s1,s2) freopen(s1,"r",stdin),freopen(s2,"w",stdout);
#define For(i,l,r) for(int i=l;i<=r;i++)
#define _For(i,l,r) for(int i=r;i>=l;i--)
using namespace std;
typedef long double lb;
typedef double db;
typedef unsigned long long ull;
typedef long long ll;
bool Begin;
const ll N=1e6+10,INF=1e18;
inline ll read(){
    ll x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')
          f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=(x<<1)+(x<<3)+(c^48);
        c=getchar();
    }
    return x*f;
}
inline void write(ll x){
	if(x<0){
		putchar('-');
		x=-x;
	}
	if(x>9)
	  write(x/10);
	putchar(x%10+'0');
}
struct Point{
    db x,y;
    friend bool operator==(const Point &a,const Point &b){
        return a.x==b.x&&a.y==b.y;
    }
    friend db dis(const Point &a,const Point &b){
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    }
}a[N];
ll n,sum;
namespace Sub1{
    db ans=INF;
    void work(){
        For(i,1,n)
          For(j,i+1,n)
            For(k,j+1,n)
              ans=min(ans,dis(a[i],a[j])+dis(a[j],a[k])+dis(a[i],a[k]));
        printf("%.10lf\n",ans);
    }
};
namespace Sub2{
    db ans=INF;
    bool cmp(Point &a,Point &b){
        return a.x<b.x;
    }
    void work(){
        sort(a+1,a+n+1,cmp);
        For(i,1,n-2)
          ans=min(ans,dis(a[i],a[i+1])+dis(a[i+1],a[i+2])+dis(a[i],a[i+2]));
        printf("%.10lf\n",ans);
    }
};
namespace Sub3{
    ll cnt=0;
    Point b[N];
    db ans=INF;
    bool cmp1(Point &a,Point &b){
        if(a.x!=b.x)
          return a.x<b.x;
        return a.y<b.y;
    }
    bool cmp2(Point &a,Point &b){
        if(a.y!=b.y)
          return a.y<b.y;
        return a.x<b.x;
    }
    db solve(ll l,ll r){
        if(l==r)
          return INF;
        ll mid=(l+r)>>1;
        ll I=a[mid].x;
        db d=min(solve(l,mid),solve(mid+1,r));
        cnt=0;
        For(i,l,r)
          if(abs(a[i].x-I)<=d/2)
            b[++cnt]=a[i];
        sort(b+1,b+cnt+1,cmp2);
        For(i,1,cnt){
            For(j,i+1,cnt){
                if(b[j].y-b[i].y>d/2)
                  break;
                For(k,j+1,cnt){
                    if(b[k].y-b[i].y>d/2)
                      break;
                    d=min(d,dis(b[i],b[j])+dis(b[i],b[k])+dis(b[j],b[k]));
                }
            }
        }
        return d;
    }
    void work(){
        sort(a+1,a+n+1,cmp1);
        printf("%.10lf\n",solve(1,n));
    }
};
int main(){
	open("A.in","A.out");
    n=read();
    For(i,1,n){
        a[i]={(db)read(),(db)read()};
        sum+=a[i].y;
    }
     if(n<=100)
       Sub1::work();
     else if(!sum)
       Sub2::work();
     else
      Sub3::work();
    return 0;
}

SP7209 CLOSEST - Closest Triplet Code：

#include<bits/stdc++.h>
#define Add(x,y) (x+y>=mod)?(x+y-mod):(x+y)
#define lowbit(x) x&(-x)
#define pi pair<ll,ll>
#define pii pair<ll,pair<ll,ll>>
#define iip pair<pair<ll,ll>,ll>
#define ppii pair<pair<ll,ll>,pair<ll,ll>>
#define fi first
#define se second
#define full(l,r,x) for(auto it=l;it!=r;it++) (*it)=x
#define Full(a) memset(a,0,sizeof(a))
#define open(s1,s2) freopen(s1,"r",stdin),freopen(s2,"w",stdout);
#define For(i,l,r) for(int i=l;i<=r;i++)
#define _For(i,l,r) for(int i=r;i>=l;i--)
using namespace std;
typedef long double lb;
typedef double db;
typedef unsigned long long ull;
typedef long long ll;
bool Begin;
const ll N=2e5+10,INF=1e18;
inline ll read(){
    ll x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')
          f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=(x<<1)+(x<<3)+(c^48);
        c=getchar();
    }
    return x*f;
}
inline void write(ll x){
	if(x<0){
		putchar('-');
		x=-x;
	}
	if(x>9)
	  write(x/10);
	putchar(x%10+'0');
}
struct Point{
    db x,y;
    friend bool operator==(const Point &a,const Point &b){
        return a.x==b.x&&a.y==b.y;
    }
    friend db dis(const Point &a,const Point &b){
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    }
}a[N];
ll n,sum;
namespace Sub1{
    db ans=INF;
    void work(){
        For(i,1,n)
          For(j,i+1,n)
            For(k,j+1,n)
              ans=min(ans,dis(a[i],a[j])+dis(a[j],a[k])+dis(a[i],a[k]));
        printf("%.3lf\n",ans);
    }
};
namespace Sub2{
    db ans=INF;
    bool cmp(Point &a,Point &b){
        return a.x<b.x;
    }
    void work(){
        sort(a+1,a+n+1,cmp);
        For(i,1,n-2)
          ans=min(ans,dis(a[i],a[i+1])+dis(a[i+1],a[i+2])+dis(a[i],a[i+2]));
        printf("%.3lf\n",ans);
    }
};
namespace Sub3{
    ll cnt=0;
    Point b[N];
    db ans=INF;
    bool cmp1(Point &a,Point &b){
        if(a.x!=b.x)
          return a.x<b.x;
        return a.y<b.y;
    }
    bool cmp2(Point &a,Point &b){
        if(a.y!=b.y)
          return a.y<b.y;
        return a.x<b.x;
    }
    db solve(ll l,ll r){
        if(l==r)
          return INF;
        ll mid=(l+r)>>1;
        ll I=a[mid].x;
        db d=min(solve(l,mid),solve(mid+1,r));
        cnt=0;
        For(i,l,r)
          if(abs(a[i].x-I)<=d/2)
            b[++cnt]=a[i];
        sort(b+1,b+cnt+1,cmp2);
        For(i,1,cnt){
            For(j,i+1,cnt){
                if(b[j].y-b[i].y>d/2)
                  break;
                For(k,j+1,cnt){
                    if(b[k].y-b[i].y>d/2)
                      break;
                    d=min(d,dis(b[i],b[j])+dis(b[i],b[k])+dis(b[j],b[k]));
                }
            }
        }
        return d;
    }
    void work(){
        sort(a+1,a+n+1,cmp1);
        printf("%.3lf\n",solve(1,n));
    }
};
int main(){
    while(1){
        n=read();
        if(n==-1)
          break;
        For(i,1,n){
            a[i]={(db)read(),(db)read()};
            sum+=a[i].y;
        }
        if(n<=100)
          Sub1::work();
        else if(!sum)
          Sub2::work();
        else
          Sub3::work();
    }
    return 0;
}