04-3. Huffman Codes (30)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue
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 a string of '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.
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
代码:
#include <stdio.h>
#include <malloc.h>
#include <string.h>
int count=0;
typedef struct HeapStruct *minheap;
typedef struct input1 *input;
struct input1
{
char a;
int weight;
input left,right;
};
struct HeapStruct
{
input element;
int size;
int Capacity;
};
//建立最小树
minheap Create(int n)
{
minheap H=malloc(sizeof(struct HeapStruct));
H->element=malloc( (n+1) * sizeof(struct input1) );
H->size = 0;
H->Capacity = n;
H->element[0].weight=0;
H->element[0].a='#';
return H;
}
//将输入元素插入
void Insert(minheap s,input p)
{
if(s->size < s->Capacity)
{
s->size++;
int parent=s->size;
int child;
int item=p->weight;
for(;;parent=child)
{
child=parent/2;
if(s->element[child].weight < item)
break;
else
s->element[parent]=s->element[child];
}
s->element[parent].weight=p->weight;
s->element[parent].a=p->a;
}
}
//now s is a minimum heap
//删除并返回最小元素
struct input1 deleteHeap(minheap s)
{
struct input1 Min,tem;
int parent,child;
if(s->size >=1)
{
Min=s->element[1];
tem=s->element[s->size--];
for(parent=1;parent*2<=s->size;parent=child)
{
child=parent*2;
if((child != s->size) && (s->element[child].weight>s->element[child+1].weight))
child++;
if(tem.weight < s->element[child].weight)
break;
else
s->element[parent]=s->element[child];
}
s->element[parent]=tem;
}
return Min;
}
//建立哈夫曼树
void HFM(minheap s)
{
int i;
struct input1 T;
int n=s->size;
for(i=1;i < n;i++)
{
struct input1 p2,p3;
p2=deleteHeap(s);
T.left=&p2;
p3=deleteHeap(s);
T.right=&p3;
T.weight=T.left->weight + T.right->weight;
printf("HFM:count previous=%d\n",count);
count=count+T.weight;
printf("HFM right and left:%d %d\n",T.left->weight,T.right->weight);
printf("HFM:count left=%d\n",count);
input p1=&T;
Insert(s,p1);
}
}
//检测一个字符串是否是另一个字符串的前缀
int equalOrElse(char *a,char *b)
{
int i=0;
while(a[i] && b[i])
{
if(a[i]==b[i])
i++;
else
break;
}
if(i==strlen(a) || i==strlen(b))
return 1;
else
return 0;
}
int main()
{
int N;
int i=0;
char q[100];
int q1[100];
printf("准备输入数据的个数:\n");
scanf("%d",&N);
getchar();
minheap s=Create(N);
printf("输入字符和权值:\n");
while(i<N)
{
scanf("%c",&q[i]);
printf("q=%c\n",q[i]);
scanf("%d",&q1[i]);
printf("q1=%d\n",q1[i]);
input p;
p=malloc(sizeof(struct input1));
p->weight=q1[i];
p->a=q[i];
p->left=NULL;
p->right=NULL;
Insert(s,p);
i++;
getchar();
}
HFM(s);
printf("wql=%d\n",count);
int M; //有几组需要检测的数据
scanf("%d",&M);
getchar();
char l[100];
char *(l1[100])[100];
int j=0,sum,z,k=0;
while(k<M)
{
sum=0;
while(j<N)
{
scanf("%c",&l[j]);
printf("l=%c\n",l[j]);
printf("j1=%d\n",j);
scanf("%s",l1[j]);
printf("l1=%s\n",l1[j]);
for(z=0;z<N;z++)
{
printf("z=%d q[]=%c\n",z,q[z]);
if(q[z]==l[j])
{
sum=sum+q1[z]*(strlen(l1[j]));
printf("sum=%d strlen=%d\n",sum,strlen(l1[j]));
break;
}
}
j++;
printf("j2=%d\n",j);
getchar();
}
if(sum==count)
{
for(i=0;i<N;i++)
{
for(j=i+1;j<N;j++)
{
if(equalOrElse(l1[i],l1[j]) == 1)
{
printf("No\n");
goto l1;
}
}
}
printf("Yes\n");
}
else
printf("No\n");
l1: k++;
j=0;
i=0;
}
return 0;
}