算法导论 15章(1)最优二叉搜索树

来源：互联网发布：成都软件招聘编辑：程序博客网时间：2024/05/20 11:26
#include <iostream>#include <vector>using namespace std;class Result{public:Result(int n){num = n+2;e = new double *[n+2];root = new int *[n+2];w = new double *[n+2];for(int i=0; i<=n+1; i++){e[i] = new double [n+2];root[i] = new int [n+2];w[i] = new double [n+2];}/****************test*******************for(i = 0; i<num; i++)for(int j=0; j<num; j++){e[i][j] = i+j;root[i][j] =i+j;}***************test********************/}void Print(){for(int i = 0; i<num; i++){for(int j = 0; j<num; j++){cout << e[i][j] << "  " ;}cout << endl;}cout << endl;for(i = 0; i<num; i++){for(int j = 0; j<num; j++){cout << root[i][j] << "  " ;}cout << endl;}}void set_w(int i, int j, double num){w[i][j] = num;}void set_e(int i, int j, double num){e[i][j] = num;}void set_root(int i, int j, int num){root[i][j] = num;}double get_w(int i, int j){return w[i][j];}double get_e(int i, int j){return e[i][j];}int get_root(int i, int j){return root[i][j];}private:double **w;   //保存概率double **e;   //保存BST的查找期望int **root;  //保存BST的节点信息int num;    //保存数组规模};Result OPTIMAL_BST(const vector<double> &p, const vector<double> &q, const int &n){//p中的有效元素只有n个，第0个元素是填充的//q中的有效元素有n+1个//这里的n指的是查找节点的个数Result res(n);for(int i = 1; i<=n+1; i++){res.set_w(i, i-1, q[i-1]);res.set_e(i, i-1, q[i-1]);}for(int l = 1; l <= n; l++)for(int i = 1; i <= n-l+1; i++){int j = i+l-1;res.set_e(i, j, 99999);res.set_w(i, j, (res.get_w(i, j-1))+p[j]+q[j]);   // w[i][j] = w[i][j-1]+p[j]+q[j]for(int r = i; r<=j; r++){double tmp = res.get_e(i, r-1)+res.get_e(r+1, j)+res.get_w(i, j);if(tmp < res.get_e(i, j)){res.set_e(i, j, tmp);res.set_root(i, j, r);}}}return res;}//打印节点的函数void Print(Result res, int i, int j, int flag){if(flag == 1){cout << "k" <<res.get_root(i, j) << "是根节点" << endl; flag = 0;}int r = res.get_root(i, j);//如果左子树是叶子if(i>r-1){//当i比r-1大时，左子树是叶子了，即为d的节点，此时d的下标就是r-1cout << "d" << r-1 << "是" << "k" << r << "的左孩子" << endl;}//如果左子树不是叶子else{cout << "k" << res.get_root(i, r-1) << "是" << "k" << r << "的左孩子" << endl;Print(res, i, r-1, 0);}//如果右子树是叶子if(j<r+1){cout << "d" << j << "是" << "k" << r << "的左孩子" << endl;}//如果左子树不是叶子else{cout << "k" << res.get_root(r+1, j) << "是" << "k" << r << "的左孩子" << endl;Print(res, r+1, j, 0);}}int main(){int n;//cin >> n;//Result res(n);vector<double> v1;vector<double> v2;v1.push_back(0);v1.push_back(0.15);v1.push_back(0.10);v1.push_back(0.05);v1.push_back(0.10);v1.push_back(0.20);v2.push_back(0.05);v2.push_back(0.10);v2.push_back(0.05);v2.push_back(0.05);v2.push_back(0.05);v2.push_back(0.10);Result res(v1.size()-1);res = OPTIMAL_BST(v1, v2, v1.size()-1);cout << res.get_e(1, 5) << endl;Print(res, 1, 5, 1);return 0;}
0 0