检查两个光盘是否在C中重叠

时间:2017-10-07 16:14:01

标签: c algorithm geometry

我必须输出非重叠光盘的数量。据说如果至少一个点重叠(它们接触),则两个盘重叠。我使用的重叠条件:
第一个输入是我应该从键盘读取的光盘数量。接下来的n行输入包含三个整数,即x,y坐标和该圆的半径 问题是我的输出不正确,我尝试了各种条件来检查光盘是否重叠,但每次我得到不同的结果。例如,当输入以下数据时,输出为5,当它应为3时。

  

10
  0 0 5
  1 7 1
  6 0 3
  -12 9 10
  8 8 6
  5 3 4
  3 2 2
  7 -10 7
  3 15 2
  -9 -7 7

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

int **readDiscs(int n) //reading discs data into array
{
    int **discs = (int **) calloc(n, sizeof(int *));

    for (int i = 0; i < n; i++)
    {
        discs[i] = (int *) calloc(3, sizeof(int));
        for (int j = 0; j < 3; j++)
            scanf("%d", &discs[i][j]);
    }
    return (discs);
}

int checkOverlap(int **discs, int length)
{
    int *overlaps = (int *) calloc(length, sizeof(int));
    int k = 0, R1, R2, X1, X2, Y1, Y2;
    for (int i = 0; i < length; i++)
        for (int j = i + 1; j < length; j++)
        {
            R1 = discs[i][2];
            R2 = discs[j][2];
            X1 = discs[i][0];
            X2 = discs[j][0];
            Y1 = discs[i][1];
            Y2 = discs[j][1];
            if (sqrt(pow(X2 - X1, 2) + pow(Y2 - Y1, 2)) <= (R1 + R2))   //if the distance is less or equal to radius,
                overlaps[i] = 1;                                        //then they overlap or at least touch
        }
    for (int f = 0; f < length; f++)
        if (overlaps[f] == 0)
            k++;
    free(overlaps);
    return (k);
}

int main(int argc, char *argv[])
{
    int **discs;
    int n;
    scanf("%d", &n);
    discs = readDiscs(n);
    printf("%d\n", checkOverlap(discs, n));
    free(discs);
    return (0);
}

1 个答案:

答案 0 :(得分:4)

您需要通过注意如果ij重叠来修复搜索,则应记录两个磁盘的搜索。因此,当重叠条件为真时,您应该有两个分配。

假设磁盘1和4(总共4个)重叠,但磁盘2和磁盘3都不与任何磁盘重叠。您记录1重叠(4),但您从未记录4重叠1。