求最近点对的基础算法

    近期算法课上，刚刚学习了 关于最近点对的相关知识，目前只是参看了大牛的思想 ，写了个最基本的裸最近点对。下面是两种方式

    
蛮力法：

#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]; 
    //数组c左半部保存划分后左部的点, 且对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); 
  //数组c左右部分分别是对y坐标有序的, 将其合并到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]; 
    //找出离划分基准左右不超过dm的部分, 且仍然对y坐标有序. 
    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); 
}