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我应该使用暴力算法来确定最近的点。我无法编译它。
算法是first algorithm on this webpage .
#include <stdio.h>
#include <float.h>
#include <stdlib.h>
#include <math.h>
struct Point
{
int x, y;
};
int compareX(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->x - p2->x);
}
int compareY(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->y - p2->y);
}
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}
float bruteForce(Point P[], int n)
{
float min = FLT_MAX;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
min = dist(P[i], P[j]);
return min;
}
float min(float x, float y)
{
return (x < y)? x : y;
}
float stripClosest(Point strip[], int size, float d)
{
float min = d; // Initialize the minimum distance as d
qsort(strip, size, sizeof(Point), compareY);
for (int i = 0; i < size; ++i)
for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
if (dist(strip[i],strip[j]) < min)
min = dist(strip[i], strip[j]);
return min;
}
float closestUtil(Point P[], int n)
{
if (n <= 3)
return bruteForce(P, n);
int mid = n/2;
Point midPoint = P[mid];
float dl = closestUtil(P, mid);
float dr = closestUtil(P + mid, n-mid);
float d = min(dl, dr);
Point strip[n];
int j = 0;
for (int i = 0; i < n; i++)
if (abs(P[i].x - midPoint.x) < d)
strip[j] = P[i], j++;
return min(d, stripClosest(strip, j, d) );
}
float closest(Point P[], int n)
{
qsort(P, n, sizeof(Point), compareX);
return closestUtil(P, n);
}
int main()
{
Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
int n = sizeof(P) / sizeof(P[0]);
printf("The smallest distance is %f ", closest(P, n));
return 0;
}
Hackerrank - Dijkstra 最短距离 2 我被卡在了 TestCase 7(我唯一失败的一个),我认为这是我的错。我下载了测试用例并检查了我生成的输出。 我执行 git diff,我看
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