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.****.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 ;
}