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
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 }