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+;
 int n;
 struct Point{
     int x, y;
 }a[N], midp[N];
 double dis(Point a, Point b){
     return sqrt(.*(a.x - b.x)*(a.x - b.x) + .*(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 +  == r) return inf;
     if(l +  == r) return dis(a[l],a[l+]) + dis(a[l+],a[r]) + dis(a[l],a[r]);

     int m = l + (r-l)/;
     double d1 = solve(l, m);
     double d2 = solve(m+, r);
     double d = min(d1, d2);
     double dd = d/2.0;
     double ans = d;

     int cnt = , i, j, k;
     for(i = l; i <= r; ++i)
         if(fabs(a[m].x - a[i].x) <= dd)
             midp[++cnt] = a[i];
     sort(midp+, midp++cnt, cmp2);

     for(i = ; i < cnt-; ++i){
         for(j = i+; j < cnt; ++j){
             if(midp[j].y - midp[i].y > dd)
                 break;
             for(k = j+; 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 = ; i <= n; ++i)
         scanf("%d%d", &a[i].x, &a[i].y);
     sort(a+, a++n, cmp1);
     printf("%.6lf\n", solve(,n));
     return ;
 }