最优二叉树<哈夫曼树>

thinking:找到2个最小值的点，将其的值加起来成一个新结点的值，然后这个新结点是这两个结点的父节点，然后再在这些结点(父结点）中选择2个接着连成一个父结点，依次循环就可以构造出一个哈夫曼树了。

the reason of failure:1、(AA&&BB||CC)与(AA&&(BB||CC))意义是不一样的。

2、得写一个walked来记录哪个结点已经被选择过了而不能再次被选择。

代码:

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
struct ttt{
	int left,right,parent,weight;
}tree[1005];

struct tt2{
	int num,weight;
};
int clock;
bool walked[1005];
int creat(int n){
	int i,j;
	tt2 min1,min2;
	int x;
	for(i=1;i<n;i++){
	memset(walked,0,sizeof(walked));
		min1.weight=9999;
		min2.weight=9999;
		for(j=1;j<=n;j++){
		x=j;
		while(tree[x].parent!=-1){
			x=tree[x].parent;
		}
			//cout << tree[x].weight << "GG" << endl;
		if(walked[x]==0&&(min1.weight>tree[x].weight||min2.weight>tree[x].weight)){
		//		cout << min1.weight << "\t" << min2.weight << endl;
				walked[x]=1;
				if(min1.weight>min2.weight){
				//	cout << "min1" << endl;
				min1.num=x;
				min1.weight=tree[x].weight;
				}else{
				//	cout << "min2" << endl;
				min2.num=x;
				min2.weight=tree[x].weight;
			}
		}
	}
	clock++;
	//cout << min1.num <<"\t" <<  min2.num << endl;
	tree[min1.num].parent=clock;
	tree[min2.num].parent=clock;
	tree[clock].weight=tree[min1.num].weight+tree[min2.num].weight;
	if(tree[min1.num].weight>tree[min1.num].weight){
		tree[clock].right=min1.num;
		tree[clock].left=min2.num;
	}else{
		tree[clock].right=min2.num;
		tree[clock].left=min1.num;
	}
	tree[clock].parent=-1;
	//cout << tree[clock].weight <<"\t" <<  min1.num<< "\t" << min2.num << endl;
}
}

int fun(int x,int t){
	if(tree[x].parent!=-1){
		fun(tree[x].parent,++t);
	}else{
		return t;
	}
}
void init(int n){
		int i;
	for(i=1;i<=n;i++){
		tree[i].parent=-1;
		tree[i].left=0;
		tree[i].right=0;
	}
}

int main(){
	freopen("in.txt","r",stdin);
	memset(tree,0,sizeof(tree));
	int i,j,k,l,n,m,t1,t2,t3,f1,f2;
	cin >> n;
	clock=n;
	for(i=1;i<=n;i++)
		cin >> tree[i].weight;
		init(n);
		creat(n);
		//print(n+n-1);
		int sum=0;
		for(i=1;i<=n;i++){
			int gg=fun(i,1);
			sum+=gg*tree[i].weight;
		}
		cout << sum << endl;
		return 0;
}