算法导论第十五章15.5-2构造最优二叉搜索树

#include "StdAfx.h"#include <iostream>  #include <vector>using namespace std;  void optimal_bst(vector<double>&p,vector<double>&q,vector<vector<double>>&e,vector<vector<double>>&w,vector<vector<int>>&root){int n = p.size() - 1;for(int i=1;i<=n+1;i++){e[i][i-1] = q[i-1];w[i][i-1] = q[i-1];}for(int l=1;l<=n;l++){for(int start = 1;start<=n-l+1;start++){int end = start + l -1; w[start][end] = w[start][end-1] + p[end] + q[end];e[start][end] = 1000;for(int r = start;r<=end;r++){double t = e[start][r-1] + e[r+1][end]  + w[start][end];if(t<e[start][end]){e[start][end] = t;root[start][end] = r;}}}}}void print(vector<vector<int>>root,int a,int b,int p){if(p == -1)printf("k%d为根\n",root[a][b]);int r = root[a][b];//左子树if(r == a){printf("d%d是k%d的左孩子\n",a-1,a);}else{printf("k%d是k%d的左孩子\n",root[a][r-1],r);print(root,a,r-1,0);}//右子树if(r == b){printf("d%d是k%d的右孩子\n",b,b);}else{printf("k%d是k%d的右孩子\n",root[r+1][b],r);print(root,r+1,b,0);}}int main(){//只需要修改这个数组便能运行不同的例子//算法导论第三版 p231 15.5-2double p1[] = {0,   0.04,0.06,0.08,0.02,0.10,0.12,0.14};double q1[] = {0.06,0.06,0.06,0.06,0.05,0.05,0.05,0.05};//算法导论第三版 p227页例子//double p1[] = {0,   0.15,0.10,0.05,0.10,0.20};//double q1[] = {0.05,0.10,0.05,0.05,0.05,0.10};vector<double>p;vector<double>q;for(int i=0;i<sizeof(p1)/sizeof(double);i++){p.push_back(p1[i]);q.push_back(q1[i]);}int n = p.size()-1;vector<vector<double>>e(n+2);vector<vector<int>>root(n+2);vector<vector<double>>w(n+2);for(int i=0;i<n+2;i++){e[i].resize(n+2);root[i].resize(n+2);w[i].resize(n+2);}optimal_bst(p,q,e,w,root);print(root,1,n,-1);return 0;}