HDU1007

分治问题（十分重要）

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67

#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <cmath>   using namespace std; struct point {     double x,y; }p[100005]; int a[100005];          // 用于记录位置号 int cmpx(const point &a,const point &b) {     return a.x < b.x; }   int cmpy(const int &a,const int &b) {     return p[a].y < p[b].y; }   double dis(int a,int b) {     return sqrt((p[a].x - p[b].x) * (p[a].x - p[b].x) + (p[a].y - p[b].y) * (p[a].y - p[b].y)); }   double cloest (int left, int right)     // 求最小距离的子函数 {     if (left == right)         return 1000000;     if (left + 1 == right)         return dis(left,right);     int mid = (left + right) >> 1;  // 相当于除二     double d1 = cloest(left, mid);     //递归求左部分最近点距离     double d2 = cloest(mid + 1, right);   //右部分     double d = min(d1, d2);     int i, j, k = 0;     for (i = left ; i <= right; i++ ) {         if (fabs(p[mid].x - p[i].x) < d)     //记录和mid位置点小于d的点的位置             a[k++] = i;        //注意这里记录的是位置号     }     sort(a, a + k, cmpy);    //按y坐标排序     for (i = 0; i < k - 1; i++) {         for (j = i + 1; j < i + 7 && j < k; j++) {             if (p[a[j]].y - p[a[i]].y >= d)                 break;             d = min(d, dis(a[i], a[j]));         }     }     return d; }     int main() {     int i, n;     while (scanf("%d", &n) != 0) {         if(!n)             break;         for( i = 0; i < n; i++ ) {             scanf("%lf %lf", &p[i].x, &p[i].y);         }         sort (p, p + n, cmpx);    //按x坐标排序         printf("%.2f\n", cloest(0, n - 1) / 2);     }     return 0; }