1004 Counting Leaves （30 分）

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

 

 

//DFS遍历方法

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;

////树遍历   
//提供每一层的儿子节点编号，给出每个结点的孩子结点
//判断每一层的叶子结点数量 
const int MAXN =100100;

vector<int>G[MAXN];//表示数 
int leaf[MAXN];//每一层的叶子结点数 
int max_h = 1;//记录深度
	 
void DFS(int index,int h){//index记录的是当前的结点编号，h是深度
	max_h = max(max_h,h);
	if(G[index].size()==0){//边界 
		leaf[h]++;
		return;
	}
	for(int i=0;i<G[index].size();i++){//size()函数获取结点数量 
		DFS(G[index][i],h+1);//枚举每一个结点； 
	}
}

int main(){
 	#ifdef ONLINE_JUDGE    
 	#else    
 		freopen("1.txt", "r", stdin);    
 	#endif 	
	int n,m,parent,child,x;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",&parent,&x);
		for(int j=0;j<x;j++){
			scanf("%d",&child);
			G[parent].push_back(child);
		}
	} 
	DFS(1,1);
	printf("%d",leaf[1]);
	for(int k=2;k<=max_h;k++)printf(" %d",leaf[k]);

	return 0;
} 

	
	

BFS遍历方法

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;

////树遍历   
//提供每一层的儿子节点
//判断每一层的叶子结点数量 
//用vector收集结点，然后遍历就OK了 
const int MAXN =100100;

vector<int>G[MAXN];//表示数 
int leaf[MAXN];//每一层的叶子结点数 
int h[MAXN];  //每个编号所在的层数 
int max_h = 1;//记录深度

void DFS(int index,int h){//////DFS遍历 ,i
	max_h = max(max_h,h);
	if(G[index].size()==0){//边界 
		leaf[h]++;
		return;
	}
	for(int i=0;i<G[index].size();i++){//size()函数获取结点数量 
		DFS(G[index][i],h+1);//枚举每一个结点； 
	}
}
void BFS(){//BFS主要是运用队列来处理 ，在这里也要注意记录最大层数 
	queue<int>q;
	q.push(1);
	while(!q.empty()){
		int id = q.front();
		q.pop();
		max_h = max(max_h,h[id]);
		if(G[id].size()==0){
			leaf[h[id]]++;
		}
		for(int i=0;i<G[id].size();i++){//这里注意孩子结点的层数变化 
			h[G[id][i]] = h[id]+1;
			q.push(G[id][i]);
		} 
		
		
	}
//	return;
	
}



int main(){
 	#ifdef ONLINE_JUDGE    
 	#else    
 		freopen("1.txt", "r", stdin);    
 	#endif 	
 	
 	fill(h,h+MAXN,0);
 	fill(leaf,leaf+MAXN,0);
	int n,m,parent,child,x;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d",&parent,&x);
		for(int j=0;j<x;j++){
			scanf("%d",&child);
			G[parent].push_back(child);
		}
	} 
	//DFS(1,1);
	h[1] = 1;//这里要注意把根结点初始化化为第一层 
	BFS();
	printf("%d",leaf[1]);
	for(int k=2;k<=max_h;k++)printf(" %d",leaf[k]);

	return 0;
}