C++ 为什么赢了';我的线段交叉测试是否返回真值?
我已经知道这个程序的输出应该是什么-我的问题是我不能让程序给出正确的输出,或者,任何输出。我的问题是:识别并显示给定巡演中跨越任何其他航段的所有航段。数据如下:旅游(“城市”)是{0,4,1,3,2},这些“城市”的积分是{2,0}、{4,1}、{0,1}、{3,2}、{1,2}。这是我的节目:C++ 为什么赢了';我的线段交叉测试是否返回真值?,c++,line,algebra,segment,C++,Line,Algebra,Segment,我已经知道这个程序的输出应该是什么-我的问题是我不能让程序给出正确的输出,或者,任何输出。我的问题是:识别并显示给定巡演中跨越任何其他航段的所有航段。数据如下:旅游(“城市”)是{0,4,1,3,2},这些“城市”的积分是{2,0}、{4,1}、{0,1}、{3,2}、{1,2}。这是我的节目: #include <iostream> using namespace std; const int MAX = 100; bool doCross(i
#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
#包括
使用名称空间std;
常数int MAX=100;
bool doCross(intx1、inty1、intx2、inty2、intx3、inty3、intx4、inty4);
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};
对于(int start=0;start您的索引看起来有点混乱。老实说,我没有在意识到您的交集算法有效后对其进行过太多调试
下面是一个简单的示例,说明了您可能正在寻找的逻辑。它的可读性当然可以通过一些临时变量来提高:
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;
}
}
}
这很有趣,因为我的教授基本上给了我们主块,而那是不起作用的部分。我很高兴是他而不是我!我将使用你的建议。非常感谢!
0 - 4 crosses 3 - 2
4 - 1 crosses 3 - 2