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：
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<stack> 5 #include<set> 6 #include<map> 7 #include<queue> 8 #include<algorithm> 9 using namespace std; 10 struct node{ 11 string dight; 12 int weight; 13 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 14 if(weight==a.weight){ 15 /*if(dight.length()==a.dight.length()){ 16 return dight.compare(a.dight)>0; 17 }*/ 18 return dight.length()<a.dight.length(); 19 } 20 return weight>a.weight; 21 } 22 }; 23 node h[100]; 24 map<char,int> ha; 25 int main(){ 26 //freopen("D:\\INPUT.txt","r",stdin); 27 int n; 28 scanf("%d",&n); 29 int i,num,sum; 30 char cha; 31 for(i=0;i<n;i++){ 32 cin>>cha; 33 scanf("%d",&ha[cha]); 34 } 35 scanf("%d",&num); 36 while(num--){ 37 sum=0; 38 priority_queue<node> q; 39 for(i=0;i<n;i++){ 40 cin>>cha>>h[i].dight; 41 //scanf("%s",h[i].dight); 42 sum+=ha[cha]; 43 h[i].weight=ha[cha]; 44 q.push(h[i]); 45 } 46 /*while(!q.empty()){ 47 cout<<q.top().weight<<" "<<q.top().dight<<endl; 48 q.pop(); 49 }*/ 50 //cout<<num<<" "<<sum<<endl; 51 node cur,next; 52 queue<node> qq; 53 bool can; 54 while(!q.empty()){ 55 cur=q.top(); 56 q.pop(); 57 can=false; 58 while(!q.empty()){ 59 next=q.top(); 60 q.pop(); 61 if(cur.dight.length()==next.dight.length()){ 62 if(cur.dight.substr(0,cur.dight.length()-1)==next.dight.substr(0,next.dight.length()-1)&&cur.dight[cur.dight.length()-1]!=next.dight[next.dight.length()-1]){ 63 can=true; 64 while(!qq.empty()){//还原 65 q.push(qq.front()); 66 qq.pop(); 67 } 68 break; 69 } 70 else{ 71 qq.push(next); 72 } 73 } 74 else{ 75 break; 76 } 77 } 78 if(can){//找到了 79 cur.dight=cur.dight.substr(0,cur.dight.length()-1); 80 cur.weight+=next.weight; 81 if(q.empty()){ 82 break; 83 } 84 q.push(cur); 85 } 86 else{ 87 break; 88 } 89 } 90 if(can&&cur.weight==sum&&!cur.dight.length()){ 91 printf("Yes\n"); 92 } 93 else{ 94 printf("No\n"); 95 } 96 } 97 return 0; 98 }
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。
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<stack> 5 #include<set> 6 #include<map> 7 #include<queue> 8 #include<algorithm> 9 using namespace std; 10 struct node{ 11 string s; 12 int count; 13 }; 14 node p[80]; 15 map<char,int> ha; 16 priority_queue<int,vector<int>,greater<int> > q;//从小到大排 17 bool check(node *p,int n){ 18 int i,j; 19 for(i=0;i<n;i++){ 20 string temp=p[i].s.substr(0,p[i].s.length()); 21 for(j=0;j<n;j++){ 22 if(i==j){ 23 continue; 24 } 25 if(temp==p[j].s.substr(0,p[i].s.length())){//前缀检查 26 break; 27 } 28 } 29 if(j<n){//不满足要求 30 return false; 31 } 32 } 33 return true; 34 } 35 int main(){ 36 //freopen("D:\\INPUT.txt","r",stdin); 37 int n,i; 38 scanf("%d",&n); 39 char c; 40 int wpl=0; 41 for(i=0;i<n;i++){ 42 cin>>c; 43 scanf("%d",&ha[c]); 44 q.push(ha[c]); 45 } 46 int cur,next; 47 while(!q.empty()){ 48 cur=q.top(); 49 q.pop(); 50 if(q.empty()){//最后一次不用做加法 51 break; 52 } 53 cur+=q.top(); 54 q.pop(); 55 wpl+=cur; 56 57 //cout<<cur<<endl; 58 59 q.push(cur); 60 } 61 //cout<<wpl<<endl; 62 int num; 63 scanf("%d",&num); 64 while(num--){ 65 int sum=0; 66 for(i=0;i<n;i++){ 67 cin>>c; 68 p[i].count=ha[c]; 69 cin>>p[i].s; 70 sum+=p[i].count*p[i].s.length(); 71 } 72 73 // cout<<sum<<endl; 74 75 if(sum==wpl&&check(p,n)){ 76 printf("Yes\n"); 77 } 78 else{ 79 printf("No\n"); 80 } 81 } 82 return 0; 83 }
