Luogu4423 BJWC2011 最小三角形 平面最近点对

传送门

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

 //This code is written by Itst
 #include<bits/stdc++.h>
 #define int long long
 #define ld long double
 #define eps (ld)1e-10
 using namespace std;

 inline int read(){
     ;
     ;
     char c = getchar();
     while(c != EOF && !isdigit(c)){
         if(c == '-')
             f = ;
         c = getchar();
     }
     while(c != EOF && isdigit(c)){
         a = (a << ) + (a << ) + (c ^ ');
         c = getchar();
     }
     return f ? -a : a;
 }

 ;
 struct node{
     int x , y;
 }now[MAXN] , pot[MAXN];
 int N;

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

 ld len(node a , node b , node c){
     return dis(a , b) + dis(b , c) + dis(a , c);
 }

 bool cmp1(node a , node b){
     return a.y < b.y;
 }

 ld solve(int l , int r){
     ){
         ld minN = 0x3f3f3f3f;
         for(int i = l ; i <= r ; i++)
              ; j <= r ; j++)
                  ; k <= r ; k++)
                     minN = min(minN , len(now[i] , now[j] , now[k]));
         return minN;
     }
     ;
     ld k = (now[mid].x + now[mid + ].x) * (ld) , r) , d = min(d1 , d2);
     sort(now + l , now + r +  , cmp1);
     ;
     for(int i = l ; i <= r ; i++)
         if(fabs(now[i].x - k) + eps < d)
             pot[++p] = now[i];
      ; i <= p ; i++)
          ; j <= p && pot[j].y - pot[i].y + eps < d ; j++)
              ; k <= p && pot[k].y - pot[i].y + eps < d ; k++)
                 d = min(d , len(pot[i] , pot[j] , pot[k]));
     return d;
 }

 bool cmp(node a , node b){
     return a.x < b.x;
 }

 signed main(){
 #ifdef LG
     freopen("4423.in" , "r" , stdin);
     //freopen("4423.out" , "w" , stdout);
 #endif
     N = read();
      ; i <= N ; i++){
         now[i].x = read();
         now[i].y = read();
     }
     sort(now +  , now + N +  , cmp);
     cout << ) << solve( , N);
     ;
 }