05-树9 Huffman Codes (30分)

题目描述

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

解题思路

根据题意，只要学生输入的WPL和通过哈夫曼树计算出来的WPL相同，并且满足前缀码的规则就输出Yes，否则输出No。于是我们先根据字符出现次数计算出最小带权路径长度，然后与学生输入比较，若相同则判断是否满足前缀码的规则。最小带权路径长度可以通过一个小顶堆计算出，而判断前缀码我使用了暴力法。

代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXSIZE 63

struct MinHeap {
    int data[MAXSIZE + 1];
    int size;
};
typedef struct MinHeap *MinHeap;

MinHeap createHeap() {
    MinHeap minHeap = (MinHeap) malloc(sizeof(struct MinHeap));
    minHeap->data[0] = 0;   //哨兵值
    minHeap->size = 0;
    return minHeap;
}

void insertHeap(MinHeap minHeap, int data) {
    if (minHeap->size == MAXSIZE) return;
    minHeap->data[++minHeap->size] = data;
    int i = minHeap->size;
    for ( ; minHeap->data[i/2] > data; i /= 2) {
        minHeap->data[i] = minHeap->data[i/2];
    }
    minHeap->data[i] = data;
}

int deleteHeap(MinHeap minHeap) {
    if (minHeap->size == 0) return 0;
    int parent, child;
    int ret = minHeap->data[1];
    minHeap->data[1] = minHeap->data[minHeap->size--];  //把最后一个元素放到堆顶
    for (parent = 1; 2 * parent <= minHeap->size; parent = child) {
        child = 2 * parent;
        //存在右儿子并且右儿子更小
        if (child < minHeap->size && minHeap->data[child+1] < minHeap->data[child]) {
            child = child + 1;
        }
        //父结点更大
        if (minHeap->data[parent] > minHeap->data[child]) {
            int tmp = minHeap->data[parent];
            minHeap->data[parent] = minHeap->data[child];
            minHeap->data[child] = tmp;
        } else {
            break;
        }
    }
    return ret;
}

//不需要构建哈夫曼树，直接算出WPL
int getWPL(MinHeap minHeap) {
    int WPL = 0, size = minHeap->size;
    int min1, min2;
    for (int i = 1; i < size; i++) {    //没有算哈夫曼树的根结点，其余结点的和刚好等于WPL，很巧妙
        min1 = deleteHeap(minHeap);
        min2 = deleteHeap(minHeap);
        insertHeap(minHeap, min1 + min2);
        WPL += min1 + min2;
    }
    return WPL;
}

//判断两个字符串是否有相同前缀
int samePrefix(char *str1, char *str2) {
    while (str1 && str2 && *str1 == *str2) {
        str1++; str2++; 
    }
    if (*str1 == ‘\0‘ || *str2 == ‘\0‘) {
        return 1;
    } else {
        return 0;
    }
}

//判断一组字符串是否满足前缀码
int isPrefixCode(char strs[][MAXSIZE], int n) {
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (samePrefix(strs[i], strs[j]))
                return 0;
        }
    }
    return 1;
}

int main() {
    int N, M, WPL, data[MAXSIZE];
    char ch;
    MinHeap minHeap = createHeap();
    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf(" %c %d", &ch, &data[i]);
        insertHeap(minHeap, data[i]);
    }
    WPL = getWPL(minHeap);
    scanf("%d", &M);
    for (int i = 0; i < M; i++) {
        int thisWPL = 0;
        char strs[MAXSIZE][MAXSIZE];
        for (int j = 0; j < N; j++) {
            scanf(" %c %s", &ch, strs[j]);
            thisWPL += data[j] * strlen(strs[j]);
        }
        if (thisWPL == WPL && isPrefixCode(strs, N)) {
            printf("Yes\n");
        } else {
            printf("No\n");
        }
    }
    return 0;
}

05-树9 Huffman Codes (30分)

原文：https://www.cnblogs.com/AndyHY-Notes/p/12569086.html