求最近點對的基礎演算法

    近期演算法課上，剛剛學習了 關於最近點對的相關知識，目前只是參看了大牛的思想 ，寫了個最基本的裸最近點對。下面是兩種方式

    
蠻力法：

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>using namespace std;struct p{   int x;   int y;};double ClosestPoint1(int n,p a[],int &index1,int &index2){       double d;       double Dist=10000;       int i,j;       for(i=0;i<n-1;i++)          for(j=i+1;j<=n-1;j++){               d=(a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y);               if(Dist>=d){                   Dist=d;                   index1=i;                   index2=j;               }          }          //printf("%d %d\n",i,j);          return Dist;}int main(){    int n,t,j;    p a[100];    scanf("%d",&n);    for(int i=0;i<n;i++){           scanf("%d%d",&a[i].x,&a[i].y);    }    int d=ClosestPoint1(n,a,t,j);   printf("%d\n",d);   return 0;  }}

分治法：

    

const int N = 100005; const double MAX = 10e100, eps = 0.00001; struct Point { double x, y; int index; }; Point a[N], b[N], c[N]; double closest(Point *, Point *, Point *, int, int); double dis(Point, Point); int cmp_x(const void *, const void*); int cmp_y(const void *, const void*); int merge(Point *, Point *, int, int, int); inline double min(double, double); int main(){     int n, i;     double d;     scanf("%d", &n);     while (n) {         for (i = 0; i < n; i++)             scanf("%lf%lf", &(a[i].x), &(a[i].y));         qsort(a, n, sizeof(a[0]), cmp_x);         for (i = 0; i < n; i++)             a[i].index = i;         memcpy(b, a, n *sizeof(a[0]));         qsort(b, n, sizeof(b[0]), cmp_y);         d = closest(a, b, c, 0, n - 1);         printf("%.2lf\n", d);         }     return 0; } double closest(Point a[],Point b[],Point c[],int p,int q){     if (q - p == 1) return dis(a[p], a[q]);     if (q - p == 2) {         double x1 = dis(a[p], a[q]);         double x2 = dis(a[p + 1], a[q]);         double x3 = dis(a[p], a[p + 1]);         if (x1 < x2 && x1 < x3) return x1;         else if (x2 < x3) return x2;         else return x3;     }     int i, j, k, m = (p + q) / 2;     double d1, d2;     for (i = p, j = p, k = m + 1; i <= q; i++)         if (b[i].index <= m) c[j++] = b[i];     //數組c左半部儲存劃分後左部的點, 且對y是有序的.     else c[k++] = b[i];     d1 = closest(a, c, b, p, m);     d2 = closest(a, c, b, m + 1, q);     double dm = min(d1, d2);   //數組c左右部分分別是對y座標有序的, 將其合并到b.     merge(b, c, p, m, q);      for (i = p, k = p; i <= q; i++)         if (fabs(b[i].x - b[m].x) < dm) c[k++] = b[i];     //找出離劃分基準左右不超過dm的部分, 且仍然對y座標有序.     for (i = p; i < k; i++)     for (j = i + 1; j < k && c[j].y - c[i].y < dm; j++){  double temp = dis(c[i], c[j]);         if (temp < dm) dm = temp;     }       return dm; } double dis(Point p, Point q){     double x1 = p.x - q.x, y1 = p.y - q.y;     return sqrt(x1 *x1 + y1 * y1); } int merge(Point p[], Point q[], int s, int m, int t){     int i, j, k;     for (i=s, j=m+1, k = s; i <= m && j <= t;) {         if (q[i].y > q[j].y) p[k++] = q[j], j++;         else p[k++] = q[i], i++;     }     while (i <= m) p[k++] = q[i++];     while (j <= t) p[k++] = q[j++];     memcpy(q + s, p + s, (t - s + 1) *sizeof(p[0]));     return 0; } int cmp_x(const void *p, const void *q){     double temp = ((Point*)p)->x - ((Point*)q)->x;     if (temp > 0) return 1;     else if (fabs(temp) < eps) return 0;     else return  - 1; } int cmp_y(const void *p, const void *q){     double temp = ((Point*)p)->y - ((Point*)q)->y;     if (temp > 0) return 1;     else if (fabs(temp) < eps) return 0;     else return  - 1; } inline double min(double p, double q) {     return (p > q) ? (q): (p); }