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c - 在此 pnpoly 算法中包括边

转载 作者:行者123 更新时间:2023-12-04 10:34:59 24 4
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我在我的寻路程序中使用多边形函数中的这个点。

int point_in_pol(int vertcount, float *vertx, float *verty, int vertexx, int vertexy){
double vertexx1;
vertexx1 = vertexx;
double vertexy1;
vertexy1 = vertexy;
int i ,j, c = 0;
for (i = 0, j = vertcount-1; i < vertcount; j = i++) {
if ( (((verty[i]>=vertexy1) && (verty[j]<=vertexy1) ) || ((verty[i]<=vertexy1) && (verty[j]>=vertexy1) )) &&
(vertexx1 < (vertx[j]-vertx[i]) * (vertexy1-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
c = !c;
}
return c;
}

如果点在多边形中,则此函数返回 true。但是,如果给定的点位于多边形的边缘,则它不会正常运行。如果该点位于边缘,我应该在此处进行哪些更改以使其返回 true?

完整代码:

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


typedef struct Node Node;
typedef struct Qnode Qnode;
void enqueue(Node* node);
void enqueue_left(Node* node);
Node* generate(int x, int y);
Node* dequeue();
void expand(Node* node, int xend, int yend);
int point_in_pol(int vertcount, float *vertx, float *verty, int vertexx, int vertexy);
struct Node{
Node* predecessor;
Node* up;
Node* down;
Node* left;
Node* right;
int xpos;
int ypos;
int marked;
};
struct Qnode{
Qnode* next;
Node* Gnode;
};
Qnode* front = NULL;
Qnode* rear = NULL;
int queuesize = 0;
int des;
int solutioncost = 0;
Node* closednodes[80000];
int nodesclosed = 0;
float polygonx[20][50];
float polygony[20][50];
int polycount = 0, vertcount = 0;
int vertcounts[20];

void enqueue(Node* node){
if (queuesize == 0){
rear = (Qnode*)malloc(sizeof(Qnode));
rear->Gnode = node;
rear->next = NULL;
front = rear;
}
else{
Qnode* temp = (Qnode*)malloc(sizeof(Qnode));
rear->next = temp;
temp->Gnode = node;
temp->next = NULL;
rear = temp;
}
queuesize++;
}
void enqueue_left(Node* node){
if(queuesize == 0){
front = (Qnode*)malloc(sizeof(Qnode));
front->Gnode = node;
front->next = NULL;
rear = front;
}
else{
Qnode* temp = (Qnode*)malloc(sizeof(Qnode));
temp->Gnode = node;
temp->next = front;
front = temp;
}
queuesize++;
}

Node* dequeue(){
Qnode* temp = front;
if (queuesize == 0){
printf("Error!");
}
else{
Node* temp1 = front->Gnode;
temp = front->next;
free(front);
front = temp;
queuesize--;
return temp1;
}

}

Node* generate(int x, int y){
int i = 0, j = 0;
//printf("Generating node (%d,%d)...\n", x, y);
if ((x >0 && x <=400) && (y >0 && y <=200)){
Node* temp = (Node*)malloc(sizeof(Node));
temp->xpos = x;
temp->ypos = y;
temp->marked = 0;
for(i; i<polycount; i++){
if(point_in_pol(vertcounts[i], polygonx[i],polygony[i], x, y)){
temp->marked = 1;
}
}
temp->up = NULL;
temp->predecessor = NULL;
temp->down = NULL;
temp->left = NULL;
temp->right = NULL;
return temp;
}
else{
printf("Invalid starting point! \n");
}
}

void expand(Node* node, int xend, int yend){
//printf("Expanding node (%d, %d)...\n", node->xpos, node->ypos);
solutioncost++;
int flag = 0;
closednodes[nodesclosed] = node;
nodesclosed++;
dequeue();
if(node->marked == 1){
// printf("Cannot expand. Part of a polygon.\n");
}

else{
if (node->xpos == xend && node->ypos == yend){
printf("Node reached!");
printf("Path in reverse order: \n(%d, %d)\n", xend, yend);
while(node->predecessor!= NULL){
printf("(%d, %d)\n", node->predecessor->xpos, node->predecessor->ypos);
node = node->predecessor;
}
queuesize = 0;
return;
}
if (node->xpos-1 >0 && node->left == NULL){
int k = 0;
int j = 0;
Qnode* temp2 = front;
for(k; k<queuesize; k++){
int xx = temp2->Gnode->xpos;
int yy = temp2->Gnode->ypos;
if (xx == node->xpos-1 && yy == node->ypos)
flag = 1;
temp2 = temp2->next;
}
for(j; j<nodesclosed; j++){
int xx = closednodes[j]->xpos;
int yy = closednodes[j]->ypos;
if (xx == node->xpos-1 && yy == node->ypos)
flag = 1;
}
if (flag == 0){
node->left = generate(node->xpos-1, node->ypos);
node->left->predecessor = node;
node->left->right = node;
if(des == 1)
enqueue(node->left);
else if(des == 2)
enqueue_left(node->left);
}
else{
//printf("(%d, %d) not generated.\n", node->xpos-1, node->ypos);
flag = 0;
}
}
if (node->xpos+1 <=400 && node->right ==NULL){
int k = 0;
int j = 0;
Qnode* temp2 = front;
for(k; k<queuesize; k++){
int xx = temp2->Gnode->xpos;
int yy = temp2->Gnode->ypos;
if (xx == node->xpos+1 && yy == node->ypos)
flag = 1;
temp2 = temp2->next;
}

for(j; j<nodesclosed; j++){
int xx = closednodes[j]->xpos;
int yy = closednodes[j]->ypos;
if (xx == node->xpos+1 && yy == node->ypos)
flag = 1;
}
if (flag == 0){
node->right = generate(node->xpos+1, node->ypos);
node->right->predecessor = node;
node->right->left = node;
if(des == 1)
enqueue(node->right);
else if(des == 2)
enqueue_left(node->right);
}
else{
//printf("(%d, %d) not generated.\n", node->xpos+1, node->ypos);
flag = 0;
}
}
if (node->ypos+1 <=200 && node->up ==NULL){
int k = 0;
int j = 0;
Qnode* temp2 = front;
for(k; k<queuesize; k++){
int xx = temp2->Gnode->xpos;
int yy = temp2->Gnode->ypos;
if (xx == node->xpos && yy == node->ypos+1)
flag = 1;
temp2 = temp2->next;
}
for(j; j<nodesclosed; j++){
int xx = closednodes[j]->xpos;
int yy = closednodes[j]->ypos;
if (xx == node->xpos && yy == node->ypos+1)
flag = 1;
}

if (flag ==0){
node->up = generate(node->xpos, node->ypos+1);
node->up->predecessor = node;
node->up->down = node;
if(des == 1)
enqueue(node->up);
else if(des == 2)
enqueue_left(node->up);
}
else{
//printf("(%d, %d) not generated.\n", node->xpos, node->ypos+1);
flag = 0;
}
}
if (node->ypos-1 >0 && node->down ==NULL){
int k = 0;
int j = 0;
Qnode* temp2 = front;
for(k; k<queuesize; k++){
int xx = temp2->Gnode->xpos;
int yy = temp2->Gnode->ypos;
if (xx == node->xpos && yy == node->ypos-1)
flag = 1;
temp2 = temp2->next;
}

for(j; j<nodesclosed; j++){
int xx = closednodes[j]->xpos;
int yy = closednodes[j]->ypos;
if (xx == node->xpos && yy == node->ypos-1)
flag = 1;
}
if (flag ==0){
node->down = generate(node->xpos, node->ypos-1);
node->down->predecessor = node;
node->down->up = node;
if(des == 1)
enqueue(node->down);
else if(des == 2)
enqueue_left(node->down);
}
else{
// printf("(%d, %d) not generated.\n", node->xpos, node->ypos-1);
flag = 0;
}
}
return;
}

}

int point_in_pol(int vertcount, float *vertx, float *verty, int vertexx, int vertexy){
double vertexx1;
vertexx1 = vertexx;
double vertexy1;
vertexy1 = vertexy;
int i ,j, c = 0;
for (i = 0, j = vertcount-1; i < vertcount; j = i++) {
if ( (((verty[i]>=vertexy1) && (verty[j]<=vertexy1) ) || ((verty[i]<=vertexy1) && (verty[j]>=vertexy1) )) &&
(vertexx1 < (vertx[j]-vertx[i]) * (vertexy1-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
c = !c;
// if ((vertexx1 == ((vertx[j]-vertx[i]) * (vertexy1-verty[i]) / (verty[j]-verty[i])+ vertx[i])) )
// return 1;
}
return c;
}

int main(){

printf("\nFILE NUMBER 1\n");

int x_start, y_start, x_end, y_end;
clock_t begin, end;
double time_spent;
FILE *fp;
fp = fopen("1.txt", "r");
if (fp == NULL)
printf("Error printing output. \n");
else
fscanf(fp, "(%d,%d)\n", &x_start, &y_start);
fscanf(fp, "(%d,%d)\n", &x_end, &y_end);
printf("Starting point is (%d, %d)\n", x_start, y_start);
printf("Ending point is (%d, %d)\n", x_end, y_end);
char temp3[255];
char* source;
int t;
while(fgets(temp3, 255, fp)!= NULL){
source = temp3;
t = 0;
printf("Polygon %d: ", polycount);
while(sscanf(source, "(%f,%f)%*[^(]%n", &polygonx[polycount][vertcount], &polygony[polycount][vertcount], &t) == 2){
printf("(%.2f,%.2f),",polygonx[polycount][vertcount], polygony[polycount][vertcount]);
source+=t;
vertcount++;
}
printf("\n");
vertcounts[polycount] = vertcount;
vertcount = 0;
polycount++;
}
printf("Select a search algorithm:\n 1. BFS\n 2: DFS\n 3: A*");
scanf("%d", &des);
if (des == 1 || des == 2){
begin = clock();
Node* start = generate(x_start, y_start);
enqueue(start);
while(queuesize!=0){
expand(front->Gnode, x_end, y_end);
}
end = clock();
time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
printf("Solution cost is: %d.\n", solutioncost);
printf("Running time is %f.\n", time_spent);

fclose(fp);
}
}

将其用作 1.txt 的输入:

(4,4)
(1,1)
(3,2),(2,2),(2,3),(3,3)

没有答案

最佳答案

int
point_in_pol(int vertcount, float *vertx, float *verty,
int vertexx, int vertexy)
{
double vertexx1;
vertexx1 = vertexx;
double vertexy1;
vertexy1 = vertexy;

int i ,j, c = 0;
for (i = 0, j = vertcount-1; i < vertcount; j = i++)
{
if ( (((verty[i]>=vertexy1) && (verty[j]<=vertexy1) )
|| ((verty[i]<=vertexy1) && (verty[j]>=vertexy1))) // this checks
// whether y-coord
// i.e. vertexy1 is
// between edge's
// vertices
&& (vertexx1 < (vertx[j]-vertx[i])
* (vertexy1-verty[i]) / (verty[j]-verty[i])
+ vertx[i]) ) // this checks
// whether x-coord
// i.e. vertexx1
// is to the left
// of the line

c = !c;
}
return c;
}

该算法的原型(prototype)名为pnpoly,其解释可见here .它的局限性之一是它无法处理恰好位于边界上的点,即它无法说明何时是这种情况。

(边界)边上的点

Pnpoly 将平面划分为多边形内部的点和多边形外部的点。边界上的点被分类为内部或外部。

我认为这应该可以解决问题:

if (vertexx1 == (vertx[j]-vertx[i]) 
* (vertexy1-verty[i]) / (verty[j]-verty[i])
+ vertx[i]) ) // this will check if vertexx1
// is equal to x-coordinate
// of what would have point on the
// edge with y=vertexy1 had
return 1;

但是你应该小心浮点错误。计算舍入误差会使结果错误。您可以添加 epsilon 比较:

if (fabs(vertexx1 - (vertx[j]-vertx[i]) 
* (vertexy1-verty[i]) / (verty[j]-verty[i])
- vertx[i]) < EPSILON)
return 1;

关于c - 在此 pnpoly 算法中包括边,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32807037/

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