1 // 2 // Created by alim on 2017/12/23. 3 // 4 5 #include<iostream> 6 #include<cmath> //求绝对值函数需要引入该头文件 7 using namespace std; 8 const int M=1000+5; 9 double c[M][M],w[M][M],p[M],q[M]; 10 int s[M][M]; 11 int n,i,j,k; 12 void Optimal_BST() 13 { 14 for(i=1;i<=n+1;i++) 15 { 16 c[i][i-1]=0.0; 17 w[i][i-1]=q[i-1]; 18 } 19 for(int t=1;t<=n;t++)//t为关键字的规模 20 //从下标为i开始的关键字到下标为j的关键字 21 for(i=1;i<=n-t+1;i++) 22 { 23 j=i+t-1; 24 w[i][j]=w[i][j-1]+p[j]+q[j]; 25 c[i][j]=c[i][i-1]+c[i+1][j];//初始化 26 s[i][j]=i;//初始化 27 //选取i+1到j之间的某个下标的关键字作为从i到j的根，如果组成的树的期望值当前最小，则k为从i到j的根节点 28 for(k=i+1;k<=j;k++) 29 { 30 double temp=c[i][k-1]+c[k+1][j]; 31 if(temp<c[i][j]&&fabs(temp-c[i][j])>1E-6)//C++中浮点数因为精度问题不可以直接比较 32 { 33 c[i][j]=temp; 34 s[i][j]=k;//k即为从下标i到j的根节点 35 } 36 } 37 c[i][j]+=w[i][j]; 38 } 39 } 40 void Construct_Optimal_BST(int i,int j,bool flag) 41 { 42 if(flag==0) 43 { 44 cout<<"S"<<s[i][j]<<" 是根"<<endl; 45 flag=1; 46 } 47 int k=s[i][j]; 48 //如果左子树是叶子 49 if(k-1<i) 50 { 51 cout<<"e"<<k-1<<" is the left child of "<<"S"<<k<<endl; 52 } 53 //如果左子树不是叶子 54 else 55 { 56 cout<<"S"<<s[i][k-1]<<" is the left child of "<<"S"<<k<<endl; 57 Construct_Optimal_BST(i,k-1,1); 58 } 59 //如果右子树是叶子 60 if(k>=j) 61 { 62 cout<<"e"<<j<<" is the right child of "<<"S"<<k<<endl; 63 } 64 //如果右子树不是叶子 65 else 66 { 67 cout<<"S"<<s[k+1][j]<<" is the right child of "<<"S"<<k<<endl; 68 Construct_Optimal_BST(k+1,j,1); 69 } 70 } 71 int main() 72 { 73 cout << "请输入关键字的个数 n:"; 74 cin >> n; 75 cout<<"请依次输入每个关键字的搜索概率:"; 76 for (i=1; i<=n; i++ ) 77 cin>>p[i]; 78 cout << "请依次输入每个虚结点的搜索概率:"; 79 for (i=0; i<=n; i++) 80 cin>>q[i]; 81 Optimal_BST(); 82 // /*用于测试 83 for(i=1; i<=n+1;i++) 84 { 85 for (j=1; j<i;j++) 86 cout <<" " ; 87 for(j=i-1;j<=n;j++) 88 cout << w[i][j]<<" " ; 89 cout << endl; 90 } 91 for(i=1; i<=n+1;i++) 92 { 93 for (j=1; j<i;j++) 94 cout <<" " ; 95 for(j=i-1; j<=n;j++) 96 cout << c[i][j]<<" " ; 97 cout << endl; 98 } 99 for(i=1; i<=n;i++) 100 { 101 for (j=1; j<i;j++) 102 cout << " " ; 103 for(j=i-1; j<=n;j++) 104 cout << s[i][j]<<" " ; 105 cout << endl; 106 } 107 cout << endl; 108 // */用于测试 109 cout<<"最小的搜索成本为:"<<c[1][n]<<endl; 110 cout<<"最优二叉搜索树为:"; 111 Construct_Optimal_BST(1,n,0); 112 return 0; 113 } 114 /** 115 * 0.04 0.09 0.08 0.02 0.12 0.14 116 * 0.06 0.08 0.10 0.07 0.05 0.05 0.10 117 */ 118 /* 119 * 请输入关键字的个数 n:6 120 6 121 请依次输入每个关键字的搜索概率:0.04 0.09 0.08 0.02 0.12 0.14 122 0.04 0.09 0.08 0.02 0.12 0.14 123 请依次输入每个虚结点的搜索概率:0.06 0.08 0.10 0.07 0.05 0.05 0.10 124 0.06 0.08 0.10 0.07 0.05 0.05 0.10 125 0.06 0.18 0.37 0.52 0.59 0.76 1 126 0.08 0.27 0.42 0.49 0.66 0.9 127 0.1 0.25 0.32 0.49 0.73 128 0.07 0.14 0.31 0.55 129 0.05 0.22 0.46 130 0.05 0.29 131 0.1 132 0 0.18 0.55 0.95 1.23 1.76 2.52 133 0 0.27 0.67 0.9 1.38 2.09 134 0 0.25 0.46 0.94 1.48 135 0 0.14 0.45 0.98 136 0 0.22 0.68 137 0 0.29 138 0 139 0 1 2 2 2 3 5 140 0 2 2 3 3 5 141 0 3 3 3 5 142 0 4 5 5 143 0 5 6 144 0 6 145 146 最小的搜索成本为:2.52 147 最优二叉搜索树为:S5 是根 148 S2 is the left child of S5 149 S1 is the left child of S2 150 e0 is the left child of S1 151 e1 is the right child of S1 152 S3 is the right child of S2 153 e2 is the left child of S3 154 S4 is the right child of S3 155 e3 is the left child of S4 156 e4 is the right child of S4 157 S6 is the right child of S5 158 e5 is the left child of S6 159 e6 is the right child of S6 160 161 * 162 */