【添加】: 创建最小堆函数

JohnDoe86 · May 28, 2018 · 7b2a568 · 7b2a568
1 parent 5a490fb
commit 7b2a568
Showing 1 changed file with 148 additions and 140 deletions.
diff --git a/eclipse/DataStructuresCode/src/05_9_Huffman_Codes.cpp b/eclipse/DataStructuresCode/src/05_9_Huffman_Codes.cpp
@@ -1,140 +1,148 @@
-/*
- * 05_9_Huffman_Codes.cpp
- *
- * Created on: 2018年5月27日
- * Author: SummerGift
- */
-/*
-
-05-树9 Huffman Codes（30 分）
-In 1953, David A. Huffman published his paper "A Method for the Construction of Minimum-Redundancy Codes",
-and hence printed his name in the history of computer science.
-As a professor who gives the final exam problem on Huffman codes,
-I am encountering a big problem: the Huffman codes are NOT unique.
-For example, given a string "aaaxuaxz", we can observe that the frequencies of the characters 'a', 'x', 'u' and 'z' are 4, 2, 1 and 1,
-respectively. We may either encode the symbols as {'a'=0, 'x'=10, 'u'=110, 'z'=111}, or in another way as {'a'=1, 'x'=01, 'u'=001, 'z'=000},
-both compress the string into 14 bits.
-Another set of code can be given as {'a'=0, 'x'=11, 'u'=100, 'z'=101},
-but {'a'=0, 'x'=01, 'u'=011, 'z'=001}is NOT correct since "aaaxuaxz" and "aazuaxax" can both be decoded from the code 00001011001001.
-The students are submitting all kinds of codes, and I need a computer program to help me determine which ones are correct and which ones are not.
-
-Input Specification:
-Each input file contains one test case. For each case,
-the first line gives an integer N (2≤N≤63), then followed by a line that contains all the N distinct characters and their frequencies in the following format:
-
-c[1] f[1] c[2] f[2] ... c[N] f[N]
-where c[i] is a character chosen from {'0' - '9', 'a' - 'z', 'A' - 'Z', '_'},
-and f[i] is the frequency of c[i] and is an integer no more than 1000.
-The next line gives a positive integer M (≤1000), then followed by M student submissions.
-Each student submission consists of N lines, each in the format:
-
-c[i] code[i]
-where c[i] is the i-th character and code[i] is an non-empty string of no more than 63 '0's and '1's.
-
-Output Specification:
-For each test case, print in each line either "Yes" if the student's submission is correct, or "No" if not.
-
-Note: The optimal solution is not necessarily generated by Huffman algorithm.
-Any prefix code with code length being optimal is considered correct.
-
-Sample Input:
-7
-A 1 B 1 C 1 D 3 E 3 F 6 G 6
-4
-A 00000
-B 00001
-C 0001
-D 001
-E 01
-F 10
-G 11
-A 01010
-B 01011
-C 0100
-D 011
-E 10
-F 11
-G 00
-A 000
-B 001
-C 010
-D 011
-E 100
-F 101
-G 110
-A 00000
-B 00001
-C 0001
-D 001
-E 00
-F 10
-G 11
-Sample Output:
-Yes
-Yes
-No
-No
-
-这道题的意思是给出一组字符和出现的频率，让我们构造他们的哈夫曼编码。
-由于哈夫曼编码不唯一，所以编码方式有多种。然后有M个同学给出了他们的编码方式，
-让我们判断他们的编码是否正确。正确输出Yes，错误输出No
-
-问题可以分成两部分:
-1.构造N个字符的哈夫曼编码，计算带权路径长度WPL，判断每个同学的编码带权路径长度是否
-等于之前计算出来的WPL.
-2.判断是否有某一个字符的编码是另一个字符的前缀码。
-
-*/
-
-#include <stdio.h>
-#include <stdlib.h>
-#include <unistd.h>
-#include <string.h>
-
-#define MAX_N 64
-#define YES 1
-#define NO 0
-#define MIN_DATA -1
-#define QUEUE_SIZE 100
-
-struct tree_node {
- int weight;
- struct tree_node* left, right;
-};
-typedef struct tree_node *tree_node_t;
-
-//构建最小堆
-struct heap_struct {
- tree_node_t data;
- int size;
- int capacity;
-};
-typedef struct heap_struct *min_heap_t;
-
-struct queue_node {
- tree_node_t data[QUEUE_SIZE];
- int rear;
- int front;
-};
-typedef struct queue_node *queue_node_t;
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+/*
+ * 05_9_Huffman_Codes.cpp
+ *
+ * Created on: 2018年5月27日
+ * Author: SummerGift
+ */
+/*
+
+05-树9 Huffman Codes（30 分）
+In 1953, David A. Huffman published his paper "A Method for the Construction of Minimum-Redundancy Codes",
+and hence printed his name in the history of computer science.
+As a professor who gives the final exam problem on Huffman codes,
+I am encountering a big problem: the Huffman codes are NOT unique.
+For example, given a string "aaaxuaxz", we can observe that the frequencies of the characters 'a', 'x', 'u' and 'z' are 4, 2, 1 and 1,
+respectively. We may either encode the symbols as {'a'=0, 'x'=10, 'u'=110, 'z'=111}, or in another way as {'a'=1, 'x'=01, 'u'=001, 'z'=000},
+both compress the string into 14 bits.
+Another set of code can be given as {'a'=0, 'x'=11, 'u'=100, 'z'=101},
+but {'a'=0, 'x'=01, 'u'=011, 'z'=001}is NOT correct since "aaaxuaxz" and "aazuaxax" can both be decoded from the code 00001011001001.
+The students are submitting all kinds of codes, and I need a computer program to help me determine which ones are correct and which ones are not.
+
+Input Specification:
+Each input file contains one test case. For each case,
+the first line gives an integer N (2≤N≤63), then followed by a line that contains all the N distinct characters and their frequencies in the following format:
+
+c[1] f[1] c[2] f[2] ... c[N] f[N]
+where c[i] is a character chosen from {'0' - '9', 'a' - 'z', 'A' - 'Z', '_'},
+and f[i] is the frequency of c[i] and is an integer no more than 1000.
+The next line gives a positive integer M (≤1000), then followed by M student submissions.
+Each student submission consists of N lines, each in the format:
+
+c[i] code[i]
+where c[i] is the i-th character and code[i] is an non-empty string of no more than 63 '0's and '1's.
+
+Output Specification:
+For each test case, print in each line either "Yes" if the student's submission is correct, or "No" if not.
+
+Note: The optimal solution is not necessarily generated by Huffman algorithm.
+Any prefix code with code length being optimal is considered correct.
+
+Sample Input:
+7
+A 1 B 1 C 1 D 3 E 3 F 6 G 6
+4
+A 00000
+B 00001
+C 0001
+D 001
+E 01
+F 10
+G 11
+A 01010
+B 01011
+C 0100
+D 011
+E 10
+F 11
+G 00
+A 000
+B 001
+C 010
+D 011
+E 100
+F 101
+G 110
+A 00000
+B 00001
+C 0001
+D 001
+E 00
+F 10
+G 11
+Sample Output:
+Yes
+Yes
+No
+No
+
+这道题的意思是给出一组字符和出现的频率，让我们构造他们的哈夫曼编码。
+由于哈夫曼编码不唯一，所以编码方式有多种。然后有M个同学给出了他们的编码方式，
+让我们判断他们的编码是否正确。正确输出Yes，错误输出No
+
+问题可以分成两部分:
+1.构造N个字符的哈夫曼编码，计算带权路径长度WPL，判断每个同学的编码带权路径长度是否
+等于之前计算出来的WPL.
+2.判断是否有某一个字符的编码是另一个字符的前缀码。
+
+*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <unistd.h>
+#include <string.h>
+
+#define MAX_N 64
+#define YES 1
+#define NO 0
+#define MIN_DATA -1
+#define QUEUE_SIZE 100
+
+struct tree_node {
+ int weight;
+ struct tree_node* left, right;
+};
+typedef struct tree_node *tree_node_t;
+
+//构建最小堆
+struct heap_struct {
+ tree_node_t data;
+ int size;
+ int capacity;
+};
+typedef struct heap_struct *min_heap_t;
+
+struct queue_node {
+ tree_node_t data[QUEUE_SIZE];
+ int rear;
+ int front;
+};
+typedef struct queue_node *queue_node_t;
+
+//创建最小堆
+tree_node_t create_min_heap(int max_size)
+{
+ min_heap_t h = (min_heap_t)malloc(sizeof(struct heap_struct));
+ h->data = malloc(sizeof(struct tree_node)*(max_size + 1));
+ h->size = 0;
+ h->capacity = max_size;
+ h->data[0].weight = MIN_DATA;
+ return h;
+}
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