luogu P1429 平面最近点对

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调半个晚上...一个板子整这么慢

之前网上一直没看明白 今天终于完事了

首先分治很容易想到 这里按x分

考虑暴力的复杂度极限 就是所有的点到中线距离相等

可以看出上下相距很远的点显然不对

那么这个很远是什么呢 是不是大于当前答案d的都可以舍去..

所以我们可以对于到中线的距离小于d的每个点找到与它y轴距离小于d的点更新答案(主要是因为画个圆太麻烦了)

然后这个可以证明每个点匹配的个数是常数 就需要用到网上流传的那个图了

至于为什么可以考虑如果一侧的d*d正方形里如果有四个以上的点那么最小值一定可以更小

代码真好写...

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 200003;
//head
typedef double D;
#define eps 1e-8
struct node {
    D x,y;
    D len() {return sqrt(x*x+y*y);}
    node operator - (const node &o) const {
        return (node){x-o.x,y-o.y};
    }
    bool operator < (const node &o) const {
        return x < o.x;
    }
}p[N];
int n;

//[l,r)
D solve(int l,int r) {
    if(r - l == 1) return 3e20;
    int m = l+r >> 1;
    D xmid = (p[m].x + p[m - 1].x) / 2.0;
    D d = min(solve(l,m),solve(m,r));
    int topb = 0,topc = 0;
    static node a[N],b[N],c[N];
    int L = l,R = m;
    int curb = 0,curc = 0;
    rep(i,l,r - 1) {
        if(L < m && (R == r || p[L].y < p[R].y) ) {
            a[i] = p[L++];
            if(xmid - a[i].x < d) b[topb++] = a[i];
        } else {
            a[i] = p[R++];
            if(a[i].x - xmid < d) c[topc++] = a[i];
        }
    }

    rep(i,l,r - 1) p[i] = a[i];
    while(curb < topb || curc < topc) {
        if(curb < topb && (curc == topc || b[curb].y < c[curc].y)) {
            per(k,curc - 1,0) {
                if(b[curb].y - c[k].y > d) break;
                d = min(d,(c[k] - b[curb]).len());
            }
            curb++;
        } else {
            per(k,curb - 1,0) {
                if(c[curc].y - b[k].y > d) break;
                d = min(d,(b[k] - c[curc]).len());
            }
            curc++;
        }
    }
    return d;
}

int main() {
    n = read();
    rep(i,1,n) scanf("%lf%lf",&p[i].x,&p[i].y);
    sort(p+1,p+n+1);
    printf("%.4f
",solve(1,n+1));
    return 0;
}