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。
1 #include<cstdio> 2 #include<cmath> 3 #include<queue> 4 #include<cstring> 5 #include<string> 6 #include<algorithm> 7 #define CLR(a,b) memset((a),(b),sizeof((a))) 8 using namespace std; 9 10 typedef long long ll; 11 const double inf = 0x3f3f3f3f; 12 const int N = 2e5+1; 13 int n; 14 struct Point{ 15 int x, y; 16 }a[N], midp[N]; 17 double dis(Point a, Point b){ 18 return sqrt(1.*(a.x - b.x)*(a.x - b.x) + 1.*(a.y - b.y)*(a.y - b.y)); 19 } 20 bool cmp1(Point a, Point b){ 21 return a.x < b.x; 22 } 23 bool cmp2(Point a, Point b){ 24 return a.y < b.y; 25 } 26 double solve(int l, int r){ 27 if(l == r ||l + 1 == r) return inf; 28 if(l + 2 == r) return dis(a[l],a[l+1]) + dis(a[l+1],a[r]) + dis(a[l],a[r]); 29 30 int m = l + (r-l)/2; 31 double d1 = solve(l, m); 32 double d2 = solve(m+1, r); 33 double d = min(d1, d2); 34 double dd = d/2.0; 35 double ans = d; 36 37 int cnt = 0, i, j, k; 38 for(i = l; i <= r; ++i) 39 if(fabs(a[m].x - a[i].x) <= dd) 40 midp[++cnt] = a[i]; 41 sort(midp+1, midp+1+cnt, cmp2); 42 43 for(i = 1; i < cnt-1; ++i){ 44 for(j = i+1; j < cnt; ++j){ 45 if(midp[j].y - midp[i].y > dd) 46 break; 47 for(k = j+1; k <= cnt; ++k){ 48 if(midp[k].y - midp[i].y > dd) 49 break; 50 double c = dis(midp[i],midp[j])+dis(midp[j],midp[k])+dis(midp[i],midp[k]); 51 ans = min(ans, c); 52 } 53 } 54 } 55 return ans; 56 } 57 int main(){ 58 scanf("%d", &n); 59 for(int i = 1; i <= n; ++i) 60 scanf("%d%d", &a[i].x, &a[i].y); 61 sort(a+1, a+1+n, cmp1); 62 printf("%.6lf ", solve(1,n)); 63 return 0; 64 }