bzoj2458: [BeiJing2011]最小三角形（分治+几何）

题目链接：bzoj2458: [BeiJing2011]最小三角形

学习推荐博客：分治法编程问题之最接近点对问题的算法分析

题解：先将所有点按x值排列，然后每次将当前区间[l,r]分成左右两半递归求解周长最小三角形。考虑到两半区间之间可能有连成最小三角形的情况，设dd为两半区间中最小三角形周长的最小值，筛选满足要求的点（x值与中点坐标x值的距离小于dd），然后按y值排序，进而暴搜出周长最小三角形。

#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;
 
typedef long long ll;
const double inf = 0x3f3f3f3f;
const int N = 2e5+1;
int n;
struct Point{
    int x, y;
}a[N], midp[N];
double dis(Point a, Point b){
    return sqrt(1.*(a.x - b.x)*(a.x - b.x) + 1.*(a.y - b.y)*(a.y - b.y));
}
bool cmp1(Point a, Point b){
    return a.x < b.x;
}
bool cmp2(Point a, Point b){
    return a.y < b.y;
}
double solve(int l, int r){
    if(l == r ||l + 1 == r) return inf;
    if(l + 2 == r) return dis(a[l],a[l+1]) + dis(a[l+1],a[r]) + dis(a[l],a[r]);
 
    int m = l + (r-l)/2;
    double d1 = solve(l, m);
    double d2 = solve(m+1, r);
    double d = min(d1, d2);
    double dd = d/2.0;
    double ans = d;
 
    int cnt = 0, i, j, k;
    for(i = l; i <= r; ++i)
        if(fabs(a[m].x - a[i].x) <= dd)
            midp[++cnt] = a[i];
    sort(midp+1, midp+1+cnt, cmp2);
 
    for(i = 1; i < cnt-1; ++i){
        for(j = i+1; j < cnt; ++j){
            if(midp[j].y - midp[i].y > dd)
                break;
            for(k = j+1; k <= cnt; ++k){
                if(midp[k].y - midp[i].y > dd)
                    break;
                double c = dis(midp[i],midp[j])+dis(midp[j],midp[k])+dis(midp[i],midp[k]);
                ans = min(ans, c);
            }
        }
    }
    return ans;
}
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d%d", &a[i].x, &a[i].y);
    sort(a+1, a+1+n, cmp1);
    printf("%.6lf\n", solve(1,n));
    return 0;
}