Luogu4423 BJWC2011 最小三角形 平面最近点对

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题意：给出$N$个点，求其中周长最小的三角形（共线的也计算在内）。$N \leq 2 \times 10^5$

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这道题唤起了我对平面最近点对的依稀记忆

考虑平面最近点对的分治，将分界线两边的求解改为求三角形的最小边长即可。

小心坐标乘积爆int

不难但就是想不出

 1 //This code is written by Itst
 2 #include<bits/stdc++.h>
 3 #define int long long
 4 #define ld long double
 5 #define eps (ld)1e-10
 6 using namespace std;
 7 
 8 inline int read(){
 9     int a = 0;
10     bool f = 0;
11     char c = getchar();
12     while(c != EOF && !isdigit(c)){
13         if(c == '-')
14             f = 1;
15         c = getchar();
16     }
17     while(c != EOF && isdigit(c)){
18         a = (a << 3) + (a << 1) + (c ^ '0');
19         c = getchar();
20     }
21     return f ? -a : a;
22 }
23 
24 const int MAXN = 200010;
25 struct node{
26     int x , y;
27 }now[MAXN] , pot[MAXN];
28 int N;
29 
30 ld dis(node a , node b){
31     return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
32 }
33 
34 ld len(node a , node b , node c){
35     return dis(a , b) + dis(b , c) + dis(a , c);
36 }
37 
38 bool cmp1(node a , node b){
39     return a.y < b.y;
40 }
41 
42 ld solve(int l , int r){
43     if(r - l <= 4){
44         ld minN = 0x3f3f3f3f;
45         for(int i = l ; i <= r ; i++)
46             for(int j = i + 1 ; j <= r ; j++)
47                 for(int k = j + 1 ; k <= r ; k++)
48                     minN = min(minN , len(now[i] , now[j] , now[k]));
49         return minN;
50     }
51     int mid = l + r >> 1;
52     ld k = (now[mid].x + now[mid + 1].x) * (ld)0.5 , d1 = solve(l , mid) , d2 = solve(mid + 1 , r) , d = min(d1 , d2);
53     sort(now + l , now + r + 1 , cmp1);
54     int p = 0;
55     for(int i = l ; i <= r ; i++)
56         if(fabs(now[i].x - k) + eps < d)
57             pot[++p] = now[i];
58     for(int i = 1 ; i <= p ; i++)
59         for(int j = i + 1 ; j <= p && pot[j].y - pot[i].y + eps < d ; j++)
60             for(int k = j + 1 ; k <= p && pot[k].y - pot[i].y + eps < d ; k++)
61                 d = min(d , len(pot[i] , pot[j] , pot[k]));
62     return d;
63 }
64 
65 bool cmp(node a , node b){
66     return a.x < b.x;
67 }
68 
69 signed main(){
70 #ifdef LG
71     freopen("4423.in" , "r" , stdin);
72     //freopen("4423.out" , "w" , stdout);
73 #endif
74     N = read();
75     for(int i = 1 ; i <= N ; i++){
76         now[i].x = read();
77         now[i].y = read();
78     }
79     sort(now + 1 , now + N + 1 , cmp);
80     cout << fixed << setprecision(6) << solve(1 , N);
81     return 0;
82 }