我有一组(X,Y)坐标,它将单位正方形划分为子矩形.假设我的坐标是 –
( x1,y1) ( x2,y2)
(0.0000,0.0000) (0.3412,0.4175)
(0.7445,0.0000) (1.0000,0.6553)
(0.7445,0.6553) (1.0000,1.0000)
(0.0000,0.6553) (0.7445,1.0000)
(0.3412,0.0000) (0.7445,0.4175)
(0.3412,0.4175) (0.7445,0.6553)
(0.0000,0.4175) (0.3412,0.6553)....etc (total 10,000 coordinates)
作为一个例子,我只采集了16组数据,这些坐标像我这样分割我的方块 –
类似框的定义
具有相似数量的邻居的那些盒子被认为是类似的盒子.对于方框[8]上方的图像,方框[13]等有4个最近邻居.所以他们被认为是类似的盒子.
下面的图片应该清楚 –
：：我的问题：：
从图像我们可以看到 –
对于方框[8],最近的方框是：
Box(1)(有4个邻居)
框[4](也有4个邻居)
盒子[14](有4个邻居)
盒子[16](有4个邻居)
因此,在这种情况下,最近框的邻居之和= 4 4 4 4 = 16
对于方框[13],最近的方框是：
框[3](有6个邻居)
框[5](也有4个邻居)
盒子[6](有3个邻居)
盒子[12](有3个邻居)
因此,最近框的邻居之和= 6 4 3 3 = 16
在这里,(类似框)框[8]和框[13] = 16 16 = 32的邻居总数.
类似地,我想将所有具有4个邻居的框分组,并找到最近框的邻居之和.并继续为每个类似的团体.
我的代码
这是我的代码.
#include <iostream> #include <cstdlib> #include <vector> #include <stdio.h> using namespace std; class Rect { public: double x1,x2,y1,y2; // coordinates Rect(double X1,double Y1,double X2,double Y2) { if (X1 < X2) { x1 = X1; x2 = X2; } else { x2 = X1; x1 = X2; } if (Y1 < Y2) { y1 = Y1; y2 = Y2; } else { y2 = Y1; y1 = Y2; } } bool isAdjacent(Rect rect) { if (x1 == rect.x1 || x1 == rect.x2 || x2 == rect.x1 || x2 == rect.x2) { // use only < when comparing y1 and rect.y2 avoids sharing only a corner if (y1 >= rect.y1 && y1 < rect.y2) { return true; } if (y2 > rect.y1 && y2 <= rect.y2) { return true; } if (rect.y1 >= y1 && rect.y1 < y2) { return true; } if (rect.y2 > y1 && rect.y2 <= y2) { return true; } } if (y1 == rect.y1 || y1 == rect.y2 || y2 == rect.y1 || y2 == rect.y2) { if (x1 >= rect.x1 && x1 < rect.x2) { return true; } if (x2 > rect.x1 && x2 <= rect.x2) { return true; } if (rect.x1 >= x1 && rect.x1 < x2) { return true; } if (rect.x2 > x1 && rect.x2 <= x2) { return true; } } return false; } }; void isNearest(int b){ vector<Rect> rects; //Rect( x1,y2 ) rects.push_back(Rect(0.0000,0.0000,0.8147,0.1355)); rects.push_back(Rect(0.8147,1.0000,0.1355)); rects.push_back(Rect(0.8147,0.1355,0.9058,0.8350)); rects.push_back(Rect(0.0000,0.1270,0.9689)); rects.push_back(Rect(0.9058,0.9134,0.2210)); rects.push_back(Rect(0.9058,0.8350,1.0000)); rects.push_back(Rect(0.8147,1.0000)); rects.push_back(Rect(0.1270,0.6324,0.3082)); rects.push_back(Rect(0.1270,0.9689,1.0000)); rects.push_back(Rect(0.0000,1.0000)); rects.push_back(Rect(0.9134,0.2210)); rects.push_back(Rect(0.9134,0.2210,0.8350)); rects.push_back(Rect(0.9058,0.8350)); rects.push_back(Rect(0.6324,0.3082)); rects.push_back(Rect(0.6324,0.3082,0.9689)); rects.push_back(Rect(0.1270,0.9689)); int nearBox_count = 0; double TotalArea=0; for (int x = 0; x < rects.size(); ++x) { if (rects[b].isAdjacent(rects[x])) { if (x==b) { continue; //this is our Box,so do not count it. } nearBox_count++; printf("Box[%d] is nearest to Box[%d] \n",(b+1),(x+1)); } } printf("Total number of nearest Box for [%d] is %d \n",nearBox_count ); printf("\n"); } int main() { for (int i = 0; i < 16; ++i) { isNearest(i); } return 0; }
它给出了正确的结果,如下所示 –
Box[1] is nearest to Box[2] Box[1] is nearest to Box[4] Box[1] is nearest to Box[8] Box[1] is nearest to Box[14] Total number of nearest Box for [1] is 4 Box[2] is nearest to Box[1] Box[2] is nearest to Box[3] Box[2] is nearest to Box[5] Box[2] is nearest to Box[11] Total number of nearest Box for [2] is 4 Box[3] is nearest to Box[2] Box[3] is nearest to Box[5] Box[3] is nearest to Box[7] Box[3] is nearest to Box[13] Box[3] is nearest to Box[14] Box[3] is nearest to Box[15] Total number of nearest Box for [3] is 6 Box[4] is nearest to Box[1] Box[4] is nearest to Box[8] Box[4] is nearest to Box[10] Box[4] is nearest to Box[16] Total number of nearest Box for [4] is 4 Box[5] is nearest to Box[2] Box[5] is nearest to Box[3] Box[5] is nearest to Box[11] Box[5] is nearest to Box[13] Total number of nearest Box for [5] is 4 Box[6] is nearest to Box[7] Box[6] is nearest to Box[12] Box[6] is nearest to Box[13] Total number of nearest Box for [6] is 3 Box[7] is nearest to Box[3] Box[7] is nearest to Box[6] Box[7] is nearest to Box[9] Box[7] is nearest to Box[15] Total number of nearest Box for [7] is 4 Box[8] is nearest to Box[1] Box[8] is nearest to Box[4] Box[8] is nearest to Box[14] Box[8] is nearest to Box[16] Total number of nearest Box for [8] is 4 Box[9] is nearest to Box[7] Box[9] is nearest to Box[10] Box[9] is nearest to Box[15] Box[9] is nearest to Box[16] Total number of nearest Box for [9] is 4 Box[10] is nearest to Box[4] Box[10] is nearest to Box[9] Total number of nearest Box for [10] is 2 Box[11] is nearest to Box[2] Box[11] is nearest to Box[5] Box[11] is nearest to Box[12] Total number of nearest Box for [11] is 3 Box[12] is nearest to Box[6] Box[12] is nearest to Box[11] Box[12] is nearest to Box[13] Total number of nearest Box for [12] is 3 Box[13] is nearest to Box[3] Box[13] is nearest to Box[5] Box[13] is nearest to Box[6] Box[13] is nearest to Box[12] Total number of nearest Box for [13] is 4 Box[14] is nearest to Box[1] Box[14] is nearest to Box[3] Box[14] is nearest to Box[8] Box[14] is nearest to Box[15] Total number of nearest Box for [14] is 4 Box[15] is nearest to Box[3] Box[15] is nearest to Box[7] Box[15] is nearest to Box[9] Box[15] is nearest to Box[14] Box[15] is nearest to Box[16] Total number of nearest Box for [15] is 5 Box[16] is nearest to Box[4] Box[16] is nearest to Box[8] Box[16] is nearest to Box[9] Box[16] is nearest to Box[15] Total number of nearest Box for [16] is 4
虽然它可以识别最近的方框并计算邻居的数量,但我无法弄清楚如何对相似的方框进行分组(如上所述)并找到总和.
我被困在这里.谁能帮我？
更新了代码段
vector<CheckRect> rects;
unsigned isNearest(unsigned b,vector<unsigned>& neighbours) {
unsigned nearBox_count = 0;
for (unsigned x = 0; x < rects.size(); ++x) {
if (rects[b].isAdjacent(rects[x])) {
if (x==b) continue; //this is our Box,so do not count it.
nearBox_count++;
printf("Box[%d] is nearest to Box[%d] \n",(x+1));
neighbours.push_back(x);
}
}
printf("Total number of nearest Box for [%d] is %d \n",nearBox_count );
printf("\n");
return nearBox_count;
}
int main(){
cin>>N;
for(int b=0; b<N; b++){
ifstream inputFile1("RectCoordinates.txt"); //input from the file prevIoUsly generated
int rect_number;
double xa0,ya0,xa1,ya1;
int neighbours;
isNearest( b,&neighbours);// This is the line that causing my ERROR
}
vector<unsigned> nearBox_count(rects.size());
vector< vector<unsigned> > neighbours(rects.size());
for (unsigned i = 0; i < rects.size(); ++i) {
nearBox_count[i] = isNearest(i,neighbours[i]);
}
// Calculate the sums of neighbouring Boxes
vector<unsigned> neighCount(rects.size(),0);
for (unsigned i = 0; i < rects.size(); i++) {
for (unsigned j = 0; j < neighbours[i].size(); j++) {
neighCount[i] += nearBox_count[neighbours[i][j]];
}
}
// Calculate your result
map<unsigned,unsigned> finalCount;
for (unsigned i = 0; i < rects.size(); i++)
{
if (finalCount.count(nearBox_count[i]) == 0)
finalCount[nearBox_count[i]] = neighCount[i];
else
finalCount[nearBox_count[i]] += neighCount[i];
}
// Print the result
for (map<unsigned,unsigned>::iterator it = finalCount.begin();
it != finalCount.end(); ++it) {
printf("Sum neighbours for the neighbours of similar Boxes with %d "
"neighbours is %d\n",it->first,it->second);
}
return 0;
}
给我错误 –
ss.cpp: In function ‘int main()’: ss.cpp:102:29: error: invalid initialization of reference of type ‘std::vector<unsigned int>&’ from expression of type ‘unsigned int’ ss.cpp:22:10: error: in passing argument 2 of ‘unsigned int isNearest(unsigned int,std::vector<unsigned int>&)’
我该如何解决？
解决方法
除了尝试计算您的价值之外,我还对您的代码进行了一些小的更改.
由于你的所有列表索引都不是负数,并且未来很可能会有大量的矩形,我建议你将所有的整数都设置为无符号.这具有额外的优点,即在下面的代码中抑制有关比较有符号和无符号整数的某些编译器警告.
我建议你做的第二个改变是只声明一次rects而不是每次迭代isNearest.在下面的代码中,我通过将rects设为全局变量并创建一个单独的函数来初始化它来实现这一点.通过将rects设为全局变量,您现在可以使用rects.size()替换所有16个(减少在添加完整数据集时忘记更改16的机会).
#include <iostream> #include <fstream> #include <cstdlib> #include <vector> #include <map> #include <stdio.h> using namespace std; class Rect { public: double x1,y2; // coordinates Rect(double X1,double Y2) { if (X1 < X2) { x1 = X1; x2 = X2; } else { x2 = X1; x1 = X2; } if (Y1 < Y2) { y1 = Y1; y2 = Y2; } else { y2 = Y1; y1 = Y2; } } bool isAdjacent(Rect rect) { if (x1 == rect.x1 || x1 == rect.x2 || x2 == rect.x1 || x2 == rect.x2) { // use only < when comparing y1 and rect.y2 avoids sharing only a corner if (y1 >= rect.y1 && y1 < rect.y2) { return true; } if (y2 > rect.y1 && y2 <= rect.y2) { return true; } if (rect.y1 >= y1 && rect.y1 < y2) { return true; } if (rect.y2 > y1 && rect.y2 <= y2) { return true; } } if (y1 == rect.y1 || y1 == rect.y2 || y2 == rect.y1 || y2 == rect.y2) { if (x1 >= rect.x1 && x1 < rect.x2) { return true; } if (x2 > rect.x1 && x2 <= rect.x2) { return true; } if (rect.x1 >= x1 && rect.x1 < x2) { return true; } if (rect.x2 > x1 && rect.x2 <= x2) { return true; } } return false; } }; vector<Rect> rects; unsigned isNearest(unsigned b,nearBox_count ); printf("\n"); return nearBox_count; } void initRects(void) { //Rect( x1,0.9689)); } void readRects(const string& filename) { ifstream fpInput(filename.c_str()); double dTemp[4]; while (true) { for (unsigned i = 0; i < 4; i++) fpInput >> dTemp[i]; if (!fpInput.good()) break; rects.push_back(Rect(dTemp[0],dTemp[1],dTemp[2],dTemp[3])); } fpInput.close(); } int main() { // Initialize the vector rects //initRects(); readRects("RectCoordinates.txt"); vector<unsigned> nearBox_count(rects.size()); vector< vector<unsigned> > neighbours(rects.size()); for (unsigned i = 0; i < rects.size(); ++i) { nearBox_count[i] = isNearest(i,unsigned> finalCount; for (unsigned i = 0; i < rects.size(); i++) { if (finalCount.count(nearBox_count[i]) == 0) { finalCount[nearBox_count[i]] = neighCount[i]; } else { finalCount[nearBox_count[i]] += neighCount[i]; } } // Print the result for (map<unsigned,it->second); } return 0; }
更新：现在可以通过在源文件中指定Rects或从外部文件加载来使用上面的代码.在上面的修改示例中,输入文件是RectCoordinates.txt：
0.0000 0.0000 0.8147 0.1355 0.8147 0.0000 1.0000 0.1355 0.8147 0.1355 0.9058 0.8350 0.0000 0.1355 0.1270 0.9689 0.9058 0.1355 0.9134 0.2210 0.9058 0.8350 1.0000 1.0000 0.8147 0.8350 0.9058 1.0000 0.1270 0.1355 0.6324 0.3082 0.1270 0.9689 0.8147 1.0000 0.0000 0.9689 0.1270 1.0000 0.9134 0.1355 1.0000 0.2210 0.9134 0.2210 1.0000 0.8350 0.9058 0.2210 0.9134 0.8350 0.6324 0.1355 0.8147 0.3082 0.6324 0.3082 0.8147 0.9689 0.1270 0.3082 0.6324 0.9689
以上输出结果：
Sum neighbours for the neighbours of similar Boxes with 2 neighbours is 8 Sum neighbours for the neighbours of similar Boxes with 3 neighbours is 32 Sum neighbours for the neighbours of similar Boxes with 4 neighbours is 165 Sum neighbours for the neighbours of similar Boxes with 5 neighbours is 22 Sum neighbours for the neighbours of similar Boxes with 6 neighbours is 25
