[英]Why won't my line segment cross test return true?
我已經知道該程序的輸出應該是什么-我的問題是我無法讓該程序給出正確的輸出,或者與此有關的任何輸出。 我的問題是:識別並顯示給定巡回路線中與其他任何路線交叉的所有路線。 數據如下:游覽(“城市”)為{0、4、1、3、2},這些“城市”的點為{2,0},{4,1},{0,1 },{3,2},{1,2}。 這是我的程序:
#include <iostream>
using namespace std;
const int MAX = 100;
bool doCross(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4);
int main()
{
int numPts = 5;
int tourAry[MAX] = {0, 4, 1, 3, 2};
int pointsAry[MAX][2] = {{2, 0},{4, 1},{0, 1},{3, 2},{1, 2}};
for(int start = 0; start <= numPts - 3; start++)
{
int startPt = tourAry[0];
int endPt = tourAry[1];
int testSegmentX1 = pointsAry[startPt][0];
int testSegmentY1 = pointsAry[startPt][1];
int testSegmentX2 = pointsAry[endPt][0];
int testSegmentY2 = pointsAry[endPt][1];
for(int nextSeg = start + 2; nextSeg <= numPts - 2; nextSeg++)
{
startPt = tourAry[2];
endPt = tourAry[3];
int startX = pointsAry[startPt][0];
int startY = pointsAry[startPt][1];
int endX = pointsAry[endPt][0];
int endY = pointsAry[endPt][1];
if(doCross(testSegmentX1, testSegmentY1, testSegmentX2, testSegmentY2, startX, startY, endX, endY))
{
cout << tourAry[start] << " - " << tourAry[start+1] << " crosses " << tourAry[nextSeg] << " - " << tourAry[nextSeg+1] << endl;
}
}//for
}//for
return 0;
}//main
bool doCross(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)
{
bool cross = true;
double denom, numerA, numerB, uA, uB;
denom = ((y4 - y3) * (x2 - x1)) - ((x4 - x3) * (y2 - y1));
numerA = ((x4 - x3) * (y1 - y3)) - ((y4 - y3) * (x1 - x3));
numerB = ((x2 - x1) * (y1 - y3)) - ((y2 - y1) * (x1 - x3));
if(denom == 0.0)
{
if(numerA == 0.0 && numerB == 0.0)
{
cross = false;
}//if
}//if
else
{
float uA = numerA / denom;
float uB = numerB / denom;
if (uA > 0.0 && uA < 1.0 && uB > 0.0 && uB < 1.0)
{
cross = true;
}//if
else
{
cross = false;
}//else
}//else
return cross;
}//doCross
我知道輸出應該是:0-4越過3-2 4-1越過3-2
任何幫助將不勝感激。
您的索引看起來有點糟了。 老實說,一旦我意識到您的交集算法可以工作,我就不會對其進行太多調試。
這是一個簡單的示例,說明您可能正在尋找的邏輯。 當然,可以使用一些臨時變量來提高其可讀性:
for(int i=0; i<numPts-1; i++) {
for(int j=i; j<numPts-1; j++) {
if(doCross(pointsAry[tourAry[i]][0], // x1
pointsAry[tourAry[i]][1], // y1
pointsAry[tourAry[i+1]][0], // x2
pointsAry[tourAry[i+1]][1], // y2
pointsAry[tourAry[j]][0], // x3
pointsAry[tourAry[j]][1], // y3
pointsAry[tourAry[j+1]][0], // x4
pointsAry[tourAry[j+1]][1])) { // y4
cout << tourAry[i] << " - " << tourAry[i+1] << " crosses " << tourAry[j] << " - " << tourAry[j+1] << endl;
}
}
}
這給出了輸出:
0 - 4 crosses 3 - 2
4 - 1 crosses 3 - 2
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