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我正在寻找一段体面的代码,以便在我的应用程序中使用其中一种算法。我找到了这个例子:http ://rosettacode.org/wiki/K-d_tree#C 但是当我把代码放在 xcode 中时,我得到一个错误,例如:

“使用未声明的标识符”,“预期的';' 在声明的最后”。我猜是缺少头文件?

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1 回答 1

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我从链接中复制了代码并做了一个小的编辑,将“swap”从一个内联嵌套函数移动到一个静态函数。使用“gcc -C99 file.c”编译,编译成功。所以,不,它不需要一些包含文件。也许你贴错了。

如果您对这个答案感到满意,您可以接受。谢谢。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>

#define MAX_DIM 3
struct kd_node_t{
    double x[MAX_DIM];
    struct kd_node_t *left, *right;
};

inline double
dist(struct kd_node_t *a, struct kd_node_t *b, int dim)
{
    double t, d = 0;
    while (dim--) {
        t = a->x[dim] - b->x[dim];
        d += t * t;
    }
    return d;
}

    static void swap(struct kd_node_t *x, struct kd_node_t *y) {
        double tmp[MAX_DIM];
        memcpy(tmp,  x->x, sizeof(tmp));
        memcpy(x->x, y->x, sizeof(tmp));
        memcpy(y->x, tmp,  sizeof(tmp));
    }

/* see quickselect method */
struct kd_node_t*
find_median(struct kd_node_t *start, struct kd_node_t *end, int idx)
{
    if (end <= start) return NULL;
    if (end == start + 1)
        return start;


    struct kd_node_t *p, *store, *md = start + (end - start) / 2;
    double pivot;
    while (1) {
        pivot = md->x[idx];

        swap(md, end - 1);
        for (store = p = start; p < end; p++) {
            if (p->x[idx] < pivot) {
                if (p != store)
                    swap(p, store);
                store++;
            }
        }
        swap(store, end - 1);

        /* median has duplicate values */
        if (store->x[idx] == md->x[idx])
            return md;

        if (store > md) end = store;
        else        start = store;
    }
}

struct kd_node_t*
make_tree(struct kd_node_t *t, int len, int i, int dim)
{
    struct kd_node_t *n;

    if (!len) return 0;

    if ((n = find_median(t, t + len, i))) {
        i = (i + 1) % dim;
        n->left  = make_tree(t, n - t, i, dim);
        n->right = make_tree(n + 1, t + len - (n + 1), i, dim);
    }
    return n;
}

/* global variable, so sue me */
int visited;

void nearest(struct kd_node_t *root, struct kd_node_t *nd, int i, int dim,
        struct kd_node_t **best, double *best_dist)
{
    double d, dx, dx2;

    if (!root) return;
    d = dist(root, nd, dim);
    dx = root->x[i] - nd->x[i];
    dx2 = dx * dx;

    visited ++;

    if (!*best || d < *best_dist) {
        *best_dist = d;
        *best = root;
    }

    /* if chance of exact match is high */
    if (!*best_dist) return;

    if (++i >= dim) i = 0;

    nearest(dx > 0 ? root->left : root->right, nd, i, dim, best, best_dist);
    if (dx2 >= *best_dist) return;
    nearest(dx > 0 ? root->right : root->left, nd, i, dim, best, best_dist);
}

#define N 1000000
#define rand1() (rand() / (double)RAND_MAX)
#define rand_pt(v) { v.x[0] = rand1(); v.x[1] = rand1(); v.x[2] = rand1(); }
int main(void)
{
    int i;
    struct kd_node_t wp[] = {
        {{2, 3}}, {{5, 4}}, {{9, 6}}, {{4, 7}}, {{8, 1}}, {{7, 2}}
    };
    struct kd_node_t this = {{9, 2}};
    struct kd_node_t *root, *found, *million;
    double best_dist;

    root = make_tree(wp, sizeof(wp) / sizeof(wp[1]), 0, 2);

    visited = 0;
    found = 0;
    nearest(root, &this, 0, 2, &found, &best_dist);

    printf(">> WP tree\nsearching for (%g, %g)\n"
        "found (%g, %g) dist %g\nseen %d nodes\n\n",
        this.x[0], this.x[1],
        found->x[0], found->x[1], sqrt(best_dist), visited);

    million = calloc(N, sizeof(struct kd_node_t));
    srand(time(0));
    for (i = 0; i < N; i++) rand_pt(million[i]);

    root = make_tree(million, N, 0, 3);
    rand_pt(this);

    visited = 0;
    found = 0;
    nearest(root, &this, 0, 3, &found, &best_dist);

    printf(">> Million tree\nsearching for (%g, %g, %g)\n"
        "found (%g, %g, %g) dist %g\nseen %d nodes\n",
        this.x[0], this.x[1], this.x[2],
        found->x[0], found->x[1], found->x[2],
        sqrt(best_dist), visited);

    /* search many random points in million tree to see average behavior.
       tree size vs avg nodes visited:
        10      ~  7
        100     ~ 16.5
        1000        ~ 25.5
        10000       ~ 32.8
        100000      ~ 38.3
        1000000     ~ 42.6
        10000000    ~ 46.7              */
    int sum = 0, test_runs = 100000;
    for (i = 0; i < test_runs; i++) {
        found = 0;
        visited = 0;
        rand_pt(this);
        nearest(root, &this, 0, 3, &found, &best_dist);
        sum += visited;
    }
    printf("\n>> Million tree\n"
        "visited %d nodes for %d random findings (%f per lookup)\n",
        sum, test_runs, sum/(double)test_runs);

    // free(million);

    return 0;
}
于 2013-09-09T15:41:50.027 回答