BZOJ 2458 最小三角形
题面
一个平面上有很多点，求他们中的点组成的周长最小的三角形的周长。
题解
跟平面最近点对差不多，也是先把区间内的点按x坐标从中间分开，递归处理，然后再处理横跨中线的三角形。
如何缩小范围？设左右两个子区间发现的最小周长是d，则与中线距离超过d / 2都没有用了，对于一个点，所有与它距离超过d / 2的点也都没有用。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < \'0\' || c > \'9\')
if(c == \'-\') op = 1;
x = c - \'0\';
while(c = getchar(), c >= \'0\' && c <= \'9\')
x = x * 10 + c - \'0\';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar(\'-\'), x = -x;
if(x >= 10) write(x / 10);
putchar(\'0\' + x % 10);
}
const int N = 200005;
int n;
struct point {
double x, y;
bool operator < (const point &b) const{
return x < b.x;
}
point operator - (const point &b){
return (point){x - b.x, y - b.y};
}
double norm(){
return sqrt(x * x + y * y);
}
} p[N], a[N], b[N], c[N];
double calc(point x, point y, point z){
return (x - y).norm() + (y - z).norm() + (z - x).norm();
}
double solve(int l, int r){
if(l == r) return 1e20;
int mid = (l + r) >> 1;
double xmid = (p[mid].x + p[mid + 1].x) / 2;
double d = min(solve(l, mid), solve(mid + 1, r));
int pos = l, pb = 0, pc = 0, pl = l, pr = mid + 1;
while(pos <= r)
if(pl <= mid && (pr > r || p[pl].y < p[pr].y)){
if(p[pl].x > xmid - d / 2) b[++pb] = p[pl];
a[pos++] = p[pl++];
}
else{
if(p[pr].x < xmid + d / 2) c[++pc] = p[pr];
a[pos++] = p[pr++];
}
for(int i = l; i <= r; i++)
p[i] = a[i];
if(l + 1 == r) return 1e20;
for(int i = 1, j = 1; i <= pb; i++){
while(j <= pc && b[i].y - c[j].y > d / 2) j++;
for(int k = j; k <= pc && abs(b[i].y - c[k].y) < d / 2; k++)
for(int h = k + 1; h <= pc && abs(b[i].y - c[h].y) < d / 2; h++)
d = min(d, calc(b[i], c[k], c[h]));
}
for(int i = 1, j = 1; i <= pc; i++){
while(j <= pb && c[i].y - b[j].y > d / 2) j++;
for(int k = j; k <= pb && abs(c[i].y - b[k].y) < d / 2; k++)
for(int h = k + 1; h <= pb && abs(c[i].y - b[h].y) < d / 2; h++)
d = min(d, calc(c[i], b[k], b[h]));
}
return d;
}
int main(){
read(n);
for(int i = 1; i <= n; i++)
scanf("%lf%lf", &p[i].x, &p[i].y);
sort(p + 1, p + n + 1);
printf("%.6lf\\n", solve(1, n));
return 0;
}