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法律研究实现融合树的 C++ 程序
融合树是一种树数据结构，它在 w 位整数上实现了一个关联数组。这是一个 C++ 程序，用于实现融合树，它在给定的二叉树上生成一个 6 位整数数组作为输入。
算法
所需函数和输入 -
Begin Take the no of elements of the tree and take the elements. Create a structure FusionTree to declare variables. Create a function init() for creating the nodes. Create a function traverse() to traverse the tree. Create a function sort() to sort the nodes of the tree. Create a function split_child() to split the nodes. Create a function insert() to insert the nodes into the tree. Create a function main(), to call insert() to create the fusion tree and then call the function traverse to display the result. End
示例代码
#include<iostream>
using namespace std;
struct FusionTree//declaration of nodes {
int *d;
FusionTree **child_ptr;
bool l;
int n;
}*r = NULL, *np = NULL, *x = NULL;
FusionTree* init()//cretae new node {
int i;
np = new FusionTree;
np->d = new int[6];
np->child_ptr = new FusionTree *[7];
np->l = true;
np->n = 0;
for (i = 0; i < 7; i++) {
np->child_ptr[i] = NULL;
}
return np;
}
void traverse(FusionTree *p)//traverse the tree {
cout<<endl;
int i;
for (i = 0; i < p->n; i++) {
if (p->l == false) {
traverse(p->child_ptr[i]);
}
cout << " " << p->d[i];
}
if (p->l == false) {
traverse(p->child_ptr[i]);
}
cout<<endl;
}
void sort(int *p, int n)//sort the tree {
int i, j, t;
for (i = 0; i < n; i++) {
for (j = i; j <= n; j++) {
if (p[i] >p[j]) {
t = p[i];
p[i] = p[j];
p[j] = t;
}
}
}
}
int split_child(FusionTree *x, int i)//split the child {
int j, mid;
FusionTree *np1, *np3, *y;
np3 = init();// initialize new node
np3->l = true;
if (i == -1) {
mid = x->d[2];//calculate mid
x->d[2] = 0;
x->n--;
np1 = init();
np1->l = false;
x->l = true;
for (j = 3; j < 6; j++) {
np3->d[j - 3] = x->d[j];
np3->child_ptr[j - 3] = x->child_ptr[j];
np3->n++;
x->d[j] = 0;
x->n--;
}
for (j = 0; j < 6; j++) {
x->child_ptr[j] = NULL;
}
np1->d[0] = mid;
np1->child_ptr[np1->n] = x;
np1->child_ptr[np1->n + 1] = np3;
np1->n++;
r = np1;
} else {
y = x->child_ptr[i];
mid = y->d[2];
y->d[2] = 0;
y->n--;
for (j = 3; j <6 ; j++) {
np3->d[j - 3] = y->d[j];
np3->n++;
y->d[j] = 0;
y->n--;
}
x->child_ptr[i + 1] = y;
x->child_ptr[i + 1] = np3;
}
return mid;
}
void insert(int a) {
int i, t;
x = r;
if (x == NULL) {
r = init();
x = r;
} else {
if (x->l== true && x->n == 6) {
t = split_child(x, -1);
x = r;
for (i = 0; i < (x->n); i++) {
if ((a >x->d[i]) && (a < x->d[i + 1])) {
i++;
break;
} else if (a < x->d[0]) {
break;
} else {
continue;
}
}
x = x->child_ptr[i];
} else {
while (x->l == false) {
for (i = 0; i < (x->n); i++) {
if ((a >x->d[i]) && (a < x->d[i + 1])) {
i++;
break;
} else if (a < x->d[0]) {
break;
} else {
continue;
}
}
if ((x->child_ptr[i])->n == 6) {
t = split_child(x, i);
x->d[x->n] = t;
x->n++;
continue;
} else {
x = x->child_ptr[i];
}
}
}
}
x->d[x->n] = a;
sort(x->d, x->n);
x->n++;
}
int main() {
int i, n, t;
cout<<"enter the no of elements to be inserted\n";
cin>>n;
for(i = 0; i < n; i++) {
cout<<"enter the element\n";
cin>>t;
insert(t);
}
cout<<"traversal of constructed fusion tree\n";
traverse(r);
}输出
enter the no of elements to be inserted 7 enter the element 10 enter the element 20 enter the element 30 enter the element 40 enter the element 50 enter the element 60 enter the element 70 traversal of constructed fusion tree 10 20 30 40 50 60 70
更新于： 30-Jul-2019
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