平面最近点对（加强版）

Posted 2021-01-27 zxyqzy

题目描述

给定平面上n个点，找出其中的一对点的距离，使得在这n个点的所有点对中，该距离为所有点对中最小的

输入输出格式

输入格式：

第一行：n；2≤n≤200000

接下来n行：每行两个实数：x y，表示一个点的行坐标和列坐标，中间用一个空格隔开。

输出格式：

仅一行，一个实数，表示最短距离，精确到小数点后面4位。

输入输出样例

输入样例#1： 复制

3
1 1
1 2
2 2

输出样例#1： 复制

1.0000

说明

0<=x,y<=10^9

就当练手了；

分治就行了；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == ‘-‘) f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}

struct node {
	int set;
	double x, y;
}a[maxn],b[maxn];

bool cmpx(node x, node y) {
	return x.x < y.x;
}

bool cmpy(node x, node y) {
	return x.y < y.y;
}

double dis(node a, node b) {
	
	return sqrt(1.0*(a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}

double sol(int l, int r) {
	if (l == r)return inf;
	int mid = (l + r) >> 1;
	double ans = inf * 1.0;
	ans = min(sol(l, mid), sol(mid + 1, r));
	int cnt = 0;
	for (int i = l; i <= r; i++) {
		if (fabs(a[i].x - a[mid].x) <= ans)b[++cnt] = a[i];
	}
	sort(b + 1, b + 1 + cnt, cmpy);
	for (int i = 1; i <= cnt; i++) {
		for (int j = i + 1; j <= cnt; j++) {
			if (b[j].y - b[i].y > ans)break;
			ans = min(ans, dis(b[i], b[j]));
		}
		
	}
	return ans;
}

int n;

int main()
{
	//ios::sync_with_stdio(0);
	rdint(n);
	for (int i = 1; i <= n; i++) {
		rdlf(a[i].x); rdlf(a[i].y); a[i].set = 0;
	}
	sort(a + 1, a + 1 + n, cmpx);
	printf("%.4lf
", sol(1, n)*1.0);
	return 0;
}