pta5-9 Huffman Codes (30分)

5-9 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

方法一：

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 #include<queue>
 #include<algorithm>
 using namespace std;
 struct node{
     string dight;
     int weight;
     bool operator<(const node &a) const {//什么情况下优先输出后面那个，这个和sort的刚好相反
         if(weight==a.weight){
             /*if(dight.length()==a.dight.length()){
                 return dight.compare(a.dight)>0;
             }*/
             return dight.length()<a.dight.length();
         }
         return weight>a.weight;
     }
 };
 node h[];
 map<char,int> ha;
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n;
     scanf("%d",&n);
     int i,num,sum;
     char cha;
     for(i=;i<n;i++){
         cin>>cha;
         scanf("%d",&ha[cha]);
     }
     scanf("%d",&num);
     while(num--){
         sum=;
         priority_queue<node> q;
         for(i=;i<n;i++){
             cin>>cha>>h[i].dight;
             //scanf("%s",h[i].dight);
             sum+=ha[cha];
             h[i].weight=ha[cha];
             q.push(h[i]);
         }
         /*while(!q.empty()){
             cout<<q.top().weight<<" "<<q.top().dight<<endl;
             q.pop();
         }*/
         //cout<<num<<" "<<sum<<endl;
         node cur,next;
         queue<node> qq;
         bool can;
         while(!q.empty()){
             cur=q.top();
             q.pop();
             can=false;
             while(!q.empty()){
                 next=q.top();
                 q.pop();
                 if(cur.dight.length()==next.dight.length()){
                    if(cur.dight.substr(,cur.dight.length()-)==next.dight.substr(,next.dight.length()-)&&cur.dight[cur.dight.length()-]!=next.dight[next.dight.length()-]){
                         can=true;
                         while(!qq.empty()){//还原
                             q.push(qq.front());
                             qq.pop();
                         }
                         break;
                    }
                    else{
                         qq.push(next);
                    }
                 }
                 else{
                     break;
                 }
             }
             if(can){//找到了
                 cur.dight=cur.dight.substr(,cur.dight.length()-);
                 cur.weight+=next.weight;
                 if(q.empty()){
                    break;
                 }
                 q.push(cur);
             }
             else{
                 break;
             }
         }
         if(can&&cur.weight==sum&&!cur.dight.length()){
             printf("Yes\n");
         }
         else{
             printf("No\n");
         }
     }
     return ;
 }

方法二：

学习网址：http://blog.csdn.net/u013167299/article/details/42244257

1.哈夫曼树法构造的wpl最小。

2.任何01字符串都不是其他字符串的前缀。

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 #include<queue>
 #include<algorithm>
 using namespace std;
 struct node{
     string s;
     int count;
 };
 node p[];
 map<char,int> ha;
 priority_queue<int,vector<int>,greater<int> > q;//从小到大排
 bool check(node *p,int n){
     int i,j;
     for(i=;i<n;i++){
         string temp=p[i].s.substr(,p[i].s.length());
         for(j=;j<n;j++){
             if(i==j){
                 continue;
             }
             if(temp==p[j].s.substr(,p[i].s.length())){//前缀检查
                 break;
             }
         }
         if(j<n){//不满足要求
             return false;
         }
     }
     return true;
 }
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,i;
     scanf("%d",&n);
     char c;
     int wpl=;
     for(i=;i<n;i++){
         cin>>c;
         scanf("%d",&ha[c]);
         q.push(ha[c]);
     }
     int cur,next;
     while(!q.empty()){
         cur=q.top();
         q.pop();
         if(q.empty()){//最后一次不用做加法
             break;
         }
         cur+=q.top();
         q.pop();
         wpl+=cur;

         //cout<<cur<<endl;

         q.push(cur);
     }
     //cout<<wpl<<endl;
     int num;
     scanf("%d",&num);
     while(num--){
         int sum=;
         for(i=;i<n;i++){
             cin>>c;
             p[i].count=ha[c];
             cin>>p[i].s;
             sum+=p[i].count*p[i].s.length();
         }

 //        cout<<sum<<endl;

         if(sum==wpl&&check(p,n)){
             printf("Yes\n");
         }
         else{
             printf("No\n");
         }
     }
     return ;
 }