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, the number of nodes in a tree, and M (<), 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

题意：
输出树中每一层的叶子节点数


方法一：层序遍历

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<cstring>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
int level[100]={0};
int n,m,id,k;
vector<vector<int> > tree;
vector<int> ans;//存放每层叶子结点数 
void findLeaf(int start){
    //层序遍历 
    queue<int> q;
    q.push(start);
    int cnt=0;
    int last=start;
    while(!q.empty()){
        int temp=q.front();//父结点 
        if(tree[temp].size()==0){
            cnt++;
        }
        for(int i=0;i<tree[temp].size();i++){
            q.push(tree[temp][i]);
        }
        if(temp==last){
                last=q.back();//把最后一个子节点赋值给last,当temp从第一个子节点到最后一个子节点时，说明该层结束 
                ans.push_back(cnt);
                cnt=0; 
        }
        q.pop();
    }
}


int main(){
    int d;
    tree.resize(100);
    scanf("%d %d",&n,&m);
    for(int i=0;i<m;i++){
        scanf("%d %d",&id,&k);
        for(int i=0;i<k;i++){
            scanf("%d",&d);
            tree[id].push_back(d);
        }
    }
    findLeaf(1);
    for(int i=0;i<ans.size();i++){
        if(i<ans.size()-1){
            printf("%d ",ans[i]);
        }
        else{
            printf("%d
",ans[i]);
        }
        
    } 
    return 0;
} 

方法二：dfs

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<cstring>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn=1010;
int n,m,id,k;
vector<vector<int> > tree;
int ans[maxn];//存放每层叶子结点数，注意此处不能用vector 
int maxlevel=-1;//保存最大层数 
void dfs(int root,int depth){
    if(tree[root].size()==0){
        if(depth>maxlevel){//若层数增加 
            maxlevel=depth; 
        }
        ans[depth]++;
        return ;
    }
    for(int i=0;i<tree[root].size();i++){
        dfs(tree[root][i],depth+1);
    }
    
}


int main(){
    int d;
    tree.resize(100);
    scanf("%d %d",&n,&m);
    for(int i=0;i<m;i++){
        scanf("%d %d",&id,&k);
        for(int i=0;i<k;i++){
            scanf("%d",&d);
            tree[id].push_back(d);
        }
    }
    dfs(1,0);
    for(int i=0;i<=maxlevel;i++){
        if(i<maxlevel){
            printf("%d ",ans[i]);
        }
        else{
            printf("%d
",ans[i]);
        }
        
    } 
    return 0;
}