05-树9 Huffman Codes 分数 30

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.

题意：

每个输入文件包含一个测试用例。对于每种情况，第一行给出一个整数N (2≤N≤63），然后后跟一行包含所有N不同的字符及其频率。下一行给出一个正整数M (≤1000），然后接着是M学生提交的材料。每个学生提交的作品包括N行

思路：

进行哈夫曼编码时用优先队列存储指向结构体的指针（或者可以用暴力每次找最小的两个）

然后先计算最短码长，与学生的进行比较，若相同则检查学生的是否是前缀码

#include <bits/stdc++.h>
using namespace std;
map<string,int>ma;		//便于计算学生的最短码长 
int ans = 0;
struct node {			//进行哈夫曼编码的node 
    string s;		
    int fre;
    node* l;
    node* r;
    int h;
};
struct CompareNodePointers {
    bool operator()(const node* a, const node* b) const {
        return a->fre > b->fre;
    }
};
struct nodet {			//进行测试学生的nodet
	string s,ss;		//s是字符，ss是学生的编码
};
node* in(node* j, node* k) {
    node* t = new node; 
    t->l = j;
    t->r = k;
    t->fre  = j->fre +k->fre ;
    return t;
}
void check(node *t){
	if(t->l && t->r ){
		ans+=t->fre ;
		check(t->l );
		check(t->r );
	}
}
int main() {
    int n,m;
    cin >> n;
    node a[n + 1];
    priority_queue<node*, vector<node*>, CompareNodePointers> q;
    for (int i = 1; i <= n; i++) {
        cin >> a[i].s >> a[i].fre;
        a[i].l  = a[i].r  = NULL;
        ma[a[i].s ] = a[i].fre ;
        node* aa = &a[i];
        q.push(aa);
    }
    node* t;
    while (q.size() > 1) {
        node* j = q.top();
        q.pop();
        node* k = q.top();
        q.pop();
        t = in(j, k);
        if(!q.empty())q.push(t);
    }
	check(t);		//计算最短码长 
	cin>>m;
	while(m--){
		int anss = 0;
		vector<nodet>test;
		for(int i = 1;i<=n;i++){
			nodet j;
			cin>>j.s >>j.ss ;
			anss+=j.ss.length() * ma[j.s]; 
			test.push_back(j);
		}
		int flag = 1;
		if(anss != ans)flag = 0;
		for (int i = 0; i < test.size(); ++i) {
        	for (int j = 0; j < test.size(); ++j) {
        		string o,p;
        		o = test[j].ss;
        		p = test[i].ss;
            	if (i != j && o.find(p) == 0) {
                	flag = 0; // 找到某个编码是另一个编码的前缀
                	break;
            	}
        	}
    	}
		if(flag)cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
    return 0;
}