04-3. Huffman Codes (30)

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本文介绍了一种验证学生提交的赫夫曼编码正确性的方法。通过计算加权路径长度(WPL)并检查编码的前缀特性来判断编码的有效性。

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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

提交代码

/*
因为有赫夫曼编码可知WPL有且只有一个最小值
虽然赫夫曼树可以有多个
还有一个条件是 短码不能是长码的前缀*/


#include <iostream>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define lson rt<<1,l,MID
#define rson rt<<1|1,MID+1,r
//#define lson root<<1
//#define rson root<<1|1
#define MID ((l+r)>>1)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=100;
const int base=1000;
const int inf=999999;
const double eps=1e-5;


struct tt
{
    char op;
    int num;
}a[maxn];


struct kk
{
    char op;
    string s;
}b[maxn];


bool cmp(kk a,kk b)
{
    return a.s.size()<b.s.size();
}


bool pan(int n)// 判断是否有短码构成了 长码的前缀码
{
    sort(b,b+n,cmp);
    for(int i=0;i<n;i++)
    {
        string tmp=b[i].s;
        for(int j=i+1;j<n;j++)
        {
            if(b[j].s.substr(0,tmp.size())==tmp)
                return false;
        }
    }
    return true;
}
int pos(char op,int n)// 返回 同学提交的字符 在原来的的字符中的位置  防止顺序不一致的出现
{
    for(int i=0;i<n;i++)
        if(a[i].op==op)return i;
        //printf("---------------------------\n");
}
int main()
{
    int n,m,i,j,k,t;
    cin>>n;
    priority_queue<int,vector<int>,greater<int> > q;
    for(i=0;i<n;i++)
    {
        cin>>a[i].op>>a[i].num;
        q.push(a[i].num);
    }
    int wpl=0;//先找到按赫夫曼编码的WPL
    while(!q.empty())
    {
        int a,b;
        a=q.top();q.pop();
        if(!q.empty()){b=q.top();q.pop();q.push(a+b);}//弹出是没有判断 是否为空  pat提示  异常退出 错了8次
        m=a+b;
        wpl+=m;
    }
    wpl-=m;
    //cout<<wpl<<endl;
    cin>>m;
    while(m--)
    {
        int wpl2=0;// 计算该同学的WPL2 和 WPL 比较
        for(i=0;i<n;i++)
        {
            cin>>b[i].op>>b[i].s;
            wpl2+=a[pos(b[i].op,n)].num*b[i].s.size();
        }
        if(wpl2>wpl)
        {
            puts("No");
        }
        else if(!pan(n))
        {
            puts("No");
        }
        else
        {
            puts("Yes");
        }
    }
    return 0;
}