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最优二叉搜索树
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StupidRabbit29
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Dec 18, 2019
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,273 @@ | ||
| #include<iostream> | ||
| #include<vector> | ||
| #include<algorithm> | ||
| #include<ctime> | ||
| #include<fstream> | ||
| #include<sstream> | ||
| using namespace std; | ||
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| #define PointX_NUM 1032 //数据个数//1032 | ||
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| class PointX | ||
| { | ||
| public: | ||
| PointX() { ID = 0; x = 0.0; y = 0.0; } | ||
| PointX(int ID, double x, double y); | ||
| ~PointX(); | ||
| int ID; | ||
| double x, y; //经度和纬度 | ||
| friend ostream& operator<<(ostream& os, const PointX& p); //重载cout | ||
| bool operator <=(PointX a)const | ||
| { | ||
| return x <= a.x; | ||
| } | ||
| bool operator <(PointX a)const | ||
| { | ||
| return x < a.x; | ||
| } | ||
| private: | ||
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| }; | ||
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| class PointY | ||
| { | ||
| public: | ||
| PointY() { ID = 0; x = 0.0; y = 0.0; } | ||
| PointY(int ID, double x, double y); | ||
| ~PointY(); | ||
| int ID; | ||
| double x, y; //经度和纬度 | ||
| friend ostream& operator<<(ostream& os, const PointY& p); //重载cout | ||
| bool operator <=(PointY a)const | ||
| { | ||
| return y <= a.y; | ||
| } | ||
| bool operator <(PointY a)const | ||
| { | ||
| return y < a.y; | ||
| } | ||
| private: | ||
|
|
||
| }; | ||
|
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| PointX::PointX(int ID, double x, double y) | ||
| { | ||
| this->ID = ID; | ||
| this->x = x; | ||
| this->y = y; | ||
| } | ||
|
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| PointX::~PointX() | ||
| { | ||
| } | ||
|
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| PointY::PointY(int ID, double x, double y) | ||
| { | ||
| this->ID = ID; | ||
| this->x = x; | ||
| this->y = y; | ||
| } | ||
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| PointY::~PointY() | ||
| { | ||
| } | ||
|
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| ostream& operator<<(ostream& os, const PointX& p) | ||
| { | ||
| cout.precision(10); // 设置输出精度 | ||
| os << "ID:" << p.ID << " (" << p.x << "," << p.y << ")"; | ||
| return os; | ||
| } | ||
| ostream& operator<<(ostream& os, const PointY& p) | ||
| { | ||
| cout.precision(10); // 设置输出精度 | ||
| os << "ID:" << p.ID << " (" << p.x << "," << p.y << ")"; | ||
| return os; | ||
| } | ||
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| bool Cpair2(PointX X[], PointY Y[], PointY Z[], int n, PointX& a, PointX& b, double& d);//平面最近点对 | ||
| void closest(PointX X[], PointY Y[], PointY Z[], int l, int r, PointX& a, PointX& b, double& d);//X以x坐标排序的点，Y以y坐标排序的点，Z存放分界线范围内的点 ,l左端点，r右端点（包含r），a,b是两个最近的点，d最短距离 | ||
| template<class Type> inline double distance(const Type& u, const Type& v); | ||
| template <class Type> void Merge(Type a[], Type b[], int l, int m, int r);//a[]待排数组，b[]目标数组,三个参数分别是数组下标left，middle，right，包含right | ||
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| int main() | ||
| { | ||
| clock_t start_time = clock(); | ||
|
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| PointX* X = new PointX[(int)PointX_NUM]; | ||
| PointY* Y = new PointY[(int)PointX_NUM]; | ||
| PointY* Z = new PointY[(int)PointX_NUM]; | ||
| ifstream fs("data2.txt", ios::in); | ||
| if (!fs) { | ||
| cout << "打开文件失败"; return 0; | ||
| } | ||
| int cnt = 0; | ||
| string lineStr; | ||
| char ch; | ||
| for (int i = 0; getline(fs, lineStr); i++) | ||
| { | ||
| stringstream ss(lineStr); | ||
| ss >> X[i].ID >> ch >> X[i].x >> ch >> X[i].y; | ||
| stringstream ss1(lineStr); | ||
| ss1 >> Y[i].ID >> ch >> Y[i].x >> ch >> Y[i].y; | ||
| } | ||
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| fs.close(); | ||
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| PointX a, b; | ||
| double d; | ||
| Cpair2(X, Y, Z, PointX_NUM, a, b, d); | ||
| cout << "最近点对是：" << endl << a << "\t" << b << endl << "最短距离是：" << d << endl; | ||
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| delete[] X; | ||
| delete[] Y; | ||
| delete[] Z; | ||
|
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| clock_t end_time = clock(); | ||
| cout << endl << "Running time is: " << static_cast<double>(end_time - start_time) / CLOCKS_PER_SEC * 1000 << "ms" << endl; | ||
| //CLOCKS_PER_SEC表示一秒钟内CPU运行的时钟周期数，用于将clock()函数的结果转化为以秒为单位的量 | ||
|
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| system("pause"); | ||
| return 0; | ||
| } | ||
|
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| //计算平面上两点间距离 | ||
| template<class Type> | ||
| inline double distance(const Type& u, const Type& v) | ||
| { | ||
| double dx = u.x - v.x; | ||
| double dy = u.y - v.y; | ||
| return sqrt(dx * dx + dy * dy); | ||
| } | ||
|
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| //归并排序 | ||
| template <class Type> | ||
| void Merge(Type a[], Type b[], int l, int m, int r) | ||
| //a[]待排数组，b[]目标数组,三个参数分别是数组下标left，middle，right，包含right | ||
| { | ||
| int i = l, j = m + 1, k = l;//i左子数组起点，j右子数组起点，k辅助数组起点 | ||
| while ((i <= m) && (j <= r)) | ||
| { | ||
| if (a[i] <= a[j]) | ||
| b[k++] = a[i++]; | ||
| else | ||
| b[k++] = a[j++]; | ||
| } | ||
| /*将剩余元素直接拷贝到b*/ | ||
| if (i > m)//左子段空了，右子段有剩余元素 | ||
| while (j <= r) | ||
| b[k++] = a[j++]; | ||
| else | ||
| while (i <= m)//左子段有剩余元素 | ||
| b[k++] = a[i++]; | ||
| } | ||
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| bool Cpair2(PointX X[], PointY Y[], PointY Z[], int n, PointX& a, PointX& b, double& d) | ||
| { | ||
| if (n < 2) return false; | ||
| sort(X, X + n); | ||
| sort(Y, Y + n); | ||
| closest(X, Y, Z, 0, n - 1, a, b, d); | ||
| return true; | ||
| } | ||
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| void closest(PointX X[], PointY Y[], PointY Z[], //X以x坐标排序的点，Y以y坐标排序的点，Z存放分界线范围内的点 | ||
| int l, int r, PointX& a, PointX& b, double& d)//l左端点，r右端点（包含r），a,b是两个最近的点，d最短距离 | ||
| { | ||
| if (r - l == 1)//两点 | ||
| { | ||
| a = X[l]; | ||
| b = X[r]; | ||
| d = distance(a, b); | ||
| return; | ||
| } | ||
| if (r - l == 2)//三点 | ||
| { | ||
| double d1 = distance(X[l], X[l + 1]); | ||
| double d2 = distance(X[l + 1], X[r]); | ||
| double d3 = distance(X[l], X[r]); | ||
| if (d1 <= d2 && d1 <= d3) | ||
| { | ||
| a = X[1]; | ||
| b = X[l + 1]; | ||
| d = d1; | ||
| return; | ||
| } | ||
| else if (d2 <= d1 && d2 <= d3) | ||
| { | ||
| a = X[l + 1]; | ||
| b = X[r]; | ||
| d = d2; | ||
| return; | ||
| } | ||
| else | ||
| { | ||
| a = X[l]; | ||
| b = X[r]; | ||
| d = d3; | ||
| return; | ||
| } | ||
| } | ||
|
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||
| /*多于三点，用分治法*/ | ||
| int m = (l + r) / 2;//m是中位数 | ||
| int f = l, g = m + 1;//f,g分别是左半段和右半段下标起点 | ||
| PointY tempP[PointX_NUM]; | ||
| int tempI = 0; | ||
| for (int i = l; i <= r; i++)//将Y[]在分界线（x中位数）左边的放在Z左半段，否则右半段 | ||
| {//Z[]存放p1到p2内按y坐标排序的点 | ||
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| if (Y[i].x < X[m].x) Z[f++] = Y[i]; | ||
| else if (Y[i].x > X[m].x) Z[g++] = Y[i]; | ||
| else//相等，先暂时放在tempI数组里，等到分完所有元素后再用tempI数组里的元素填充满数组，使两个数组大小一样。 | ||
| { | ||
| tempP[tempI++] = Y[i]; | ||
| } | ||
| } | ||
| for (int i = 0; i < tempI; i++) | ||
| { | ||
| if (f <= m)//左边还没满，放左边 | ||
| Z[f++] = tempP[i]; | ||
| else if (g <= r)//左边满了右边没满，放右边 | ||
| Z[g++] = tempP[i]; | ||
| else | ||
| cerr << "Z[] array error"; | ||
| } | ||
|
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| closest(X, Z, Y, l, m, a, b, d);//求左半段最小距离 | ||
| double dr;//右边最短距离 | ||
| PointX ar, br; | ||
| closest(X, Z, Y, m + 1, r, ar, br, dr); | ||
| if (dr < d)//取左段和右段较小者 | ||
| { | ||
| a = ar; | ||
| b = br; | ||
| d = dr; | ||
| } | ||
|
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| Merge(Z, Y, l, m, r);//将Z里的两段数组合并放回Y | ||
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| int k = l; | ||
| for (int i = l; i <= r; i++)//将分界线2d范围内的点拷到Z数组 | ||
| { | ||
| if (fabs(Y[m].x - Y[i].x <= d)) //好像可以取等？ | ||
| Z[k++] = Y[i]; | ||
| } | ||
| for (int i = l; i < k; i++)//遍历分界线区域内的点，计算距离//不能取等 | ||
| { | ||
| for (int j = i + 1; j < k && Z[j].y - Z[i].y < d; j++)//对于点Z[i]，计算其上距离在d内的点于它的距离 | ||
| { | ||
| double dp = distance(Z[j], Z[i]); | ||
| if (dp < d)//取两段和中间距离较小者 | ||
| { | ||
| d = dp; | ||
| PointX tempA(Z[i].ID, Z[i].x, Z[i].y), tempB(Z[j].ID, Z[j].x, Z[j].y); | ||
| a = tempA; | ||
| b = tempB; | ||
| } | ||
| } | ||
| } | ||
| } | ||
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