函数SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2)是从1到upbound中找出father为0的节点赋给s1,s2,(为了保证答案唯一,请让s1的节点编号小于s2),函数HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n)是构造哈夫曼树以及计算哈夫曼编码。保证输入的权重值小于1000。
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2);
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n);
其中 upbound
编号,HT
是哈夫曼树,HC
是哈夫曼编码,w
是权值,n
是叶子节点个数。
#include
#include
#include
typedef struct {
int weight;
int parent;
int lchild;
int rchild;
} HTNode, *HuffmanTree;
typedef char ** HuffmanCode;
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2);
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n);
int main() {
HuffmanTree ht;
HuffmanCode hc;
int n;
scanf("%d", &n);
int *w = (int *) malloc (n * sizeof(int));
for(int i = 0; i < n; ++ i)
scanf("%d", &w[i]);
HuffmanCoding(ht, hc, w, n);
for (int i = 1; i <= 2 * n - 1; ++ i) {
printf("%d %d %d %d\n",
ht[i].weight, ht[i].parent, ht[i].lchild, ht[i].rchild);
}
for (int i = 1; i <= n; ++ i)
printf("%s\n", hc[i]);
free(w);
free(ht);
for (int i = 1; i <= n; ++ i)
free(hc[i]);
return 0;
}
/* 你的代码将被嵌在这里 */
####输入格式:
第一行输入一个数n,表示叶子节点的个数,接下去输入n个整数,表示每个节点的值
####输出格式:
只要建树即可,输出已经确定了
4
1 2 3 4
1 5 0 0
2 5 0 0
3 6 0 0
4 7 0 0
3 6 1 2
6 7 3 5
10 0 4 6
110
111
10
0
代码:
#include
#include
#include
typedef struct {
int weight;
int parent;
int lchild;
int rchild;
} HTNode, *HuffmanTree;
typedef char ** HuffmanCode;
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2);
void HuffmanCoding(HuffmanTree &HT, HuffmanCode &HC, int *w, int n);
int main() {
HuffmanTree ht;
HuffmanCode hc;
int n;
scanf("%d", &n);
int *w = (int *) malloc (n * sizeof(int));
for(int i = 0; i < n; ++ i)
scanf("%d", &w[i]);
HuffmanCoding(ht, hc, w, n);
for (int i = 1; i <= 2 * n - 1; ++ i) {
printf("%d %d %d %d\n",
ht[i].weight, ht[i].parent, ht[i].lchild, ht[i].rchild);
}
for (int i = 1; i <= n; ++ i)
printf("%s\n", hc[i]);
free(w);
free(ht);
for (int i = 1; i <= n; ++ i)
free(hc[i]);
return 0;
}
/* 你的代码将被嵌在这里 */
void SelectTwoMin(int upbound, HuffmanTree HT, int &s1, int &s2)
{
int min1=10000,min2=10000,num1,num2;
for(int i=1;i<=upbound;i++)
{
if(HT[i].parent==0&&HT[i].weight