//最小堆求哈弗曼編碼
#include<stdio.h>
#include<iostream>
#include<queue> 
#include<string>
using namespace std;
#define MAX 100


struct Node{
	int weight;			//權重 
	char c;				//字元 
	string code;		     //解碼結果 
	Node *parent,*left,*right;
	
	//初始化 
	Node(){
		parent  = left = right = NULL;
		code=""; 
	}
};

//定義指標比較函式 
struct PCmp
{
    bool operator () (Node const *x, Node const *y)
    {
    	 return x->weight > y->weight;
    }
}; 

priority_queue<Node *,vector<Node *>,PCmp>Q;	//最小堆
 
//構造HuffmanTree 
void CreatHuffmanTree(Node *a[],int n){
	for(int i=0;i<n;i++){
		Q.push(a[i]);
	}	
	Node *parent;
	Node *s1,*s2;
	
	while(Q.empty()==false){
		s1 = Q.top();
		Q.pop();
		if(Q.empty()==true) break;		//最後佇列中只有一個元素，也就是已經生成HuffmanTree 
		s2 = Q.top();
		Q.pop();
		
		parent=new Node();
		parent->c =' ';
		parent->weight = s1->weight+s2->weight; //設定權重 
		
		//設定子節點 
		parent->right = s2;
		parent->left = s1; 
		
		//設定父節點 
		s1->parent = parent;
		s2->parent = parent;
		
		Q.push(parent);
	}
}

//Huffman解碼 
void HuffmanCode(Node *a[],int n){
	for(int i=0;i<n;i++){
		Node *cur = a[i];
		Node *parent = a[i]->parent;
		string code="";
		
		while(parent!=NULL){	//到達根節點的時候解碼結束 
			if(parent->left==cur) code = "0"+code;
			else code = "1"+code;
			
			//結點向上一層 
			cur = parent;
			parent = parent->parent;
		}
		a[i]->code = code;//賦值編碼 
	}
} 

int main(){
	Node *initNode[MAX] ;//儲存葉子節點資訊 
	
	//freopen("HuffmanTreeData.txt","r",stdin); 
	int n;
	scanf("%d",&n);
	//輸入 資訊 
	for(int i=0;i<n;i++){
		Node *newNode = new Node();
		int weight ;
		char c;
		scanf(" %c%d",&c,&weight);
		newNode->c = c;
		newNode->weight = weight;
		initNode[i] = newNode;
	}
	
	CreatHuffmanTree(initNode,n);
	HuffmanCode(initNode,n);
	
	//輸出資訊 
	for(int i=0;i<n;i++){
		cout<<initNode[i]->c<<" "<<initNode[i]->code<<endl;
	}
	return 0;
}
//測試資料
6
a 45
b 13
c 12
d 16
e 9
f 5
//結果
a  0
b  101
c  100
d  111
e  1101
f   1100


//*下面是建樹求哈弗曼編碼*//
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

#define N 10         // 帶編碼字元的個數，即樹中葉結點的最大個數
#define M (2*N-1)    // 樹中總的結點數目

class HTNode{        // 樹中結點的結構
public: 
	unsigned int weight;
	unsigned int parent,lchild,rchild;
};                    

class HTCode{
public:
	char data;      // 待編碼的字元
	int weight;     // 字元的權值
	char code[N];   // 字元的編碼
};

void Init(HTCode hc[], int *n){
// 初始化，讀入待編碼字元的個數n，從鍵盤輸入n個字元和n個權值
	int i;
	printf("input n = ");
	scanf("%d",&(*n));

	printf("\ninput %d character\n",*n);
 	
	fflush(stdin);
	for(i=1; i<=*n; ++i)
		scanf("%c",&hc[i].data);

	printf("\ninput %d weight\n",*n);
	
	for(i=1; i<=*n; ++i)
		scanf("%d",&(hc[i].weight) );
	fflush(stdin);
}//

void Select(HTNode ht[], int k, int *s1, int *s2){
// ht[1...k]中選擇parent為0，並且weight最小的兩個結點，其序號由指標變數s1，s2指示
	int i;
	for(i=1; i<=k && ht[i].parent != 0; ++i){ 
		; ;
	}
	*s1 = i;

	for(i=1; i<=k; ++i){
		if(ht[i].parent==0 && ht[i].weight<ht[*s1].weight)
			*s1 = i;
	}

	for(i=1; i<=k; ++i){
		if(ht[i].parent==0 && i!=*s1)
			break;
	}
	*s2 = i;

	for(i=1; i<=k; ++i){
		if(ht[i].parent==0 && i!=*s1 && ht[i].weight<ht[*s2].weight)
			*s2 = i;
	}
}

void HuffmanCoding(HTNode ht[],HTCode hc[],int n){
// 構造Huffman樹ht，並求出n個字元的編碼
	char cd[N];
	int i,j,m,c,f,s1,s2,start;
	m = 2*n-1;
	
	for(i=1; i<=m; ++i){
		if(i <= n)
			ht[i].weight = hc[i].weight;
		else
			ht[i].parent = 0;
		ht[i].parent = ht[i].lchild = ht[i].rchild = 0;
	}

	for(i=n+1; i<=m; ++i){
		Select(ht, i-1, &s1, &s2);
		ht[s1].parent = i;
		ht[s2].parent = i;
		ht[i].lchild = s1;
		ht[i].rchild = s2;
		ht[i].weight = ht[s1].weight+ht[s2].weight;
	}

	cd[n-1] = '\0';

	for(i=1; i<=n; ++i){
		start = n-1;
		for(c=i,f=ht[i].parent; f; c=f,f=ht[f].parent){
			if(ht[f].lchild == c)
				cd[--start] = '0';
			else
				cd[--start] = '1';
		}
		strcpy(hc[i].code, &cd[start]);
	}
}


int main()
{
	int i,m,n,w[N+1];
	HTNode ht[M+1];
	HTCode hc[N+1];
	Init(hc, &n);     // 初始化
 	HuffmanCoding(ht,hc,n);   // 構造Huffman樹，並形成字元的編碼

	for(i=1; i<=n; ++i)  
		printf("\n%c---%s",hc[i].data,hc[i].code);  
	printf("\n");

	return 0;
}

//測試資料
6
abcdef 
45
13 
12
16 
9
5
//結果
a---0
b---101
c---100
d---111
e---1101
f ---1100