1004. Counting Leaves (30)

1004. Counting Leaves (30)

 

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

Input

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.

Output

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
 
有关树的遍历的问题，用到了深搜的方法
树的存放很有意思，map<int,vector<int> >;  key为某个节点的ID，vector是他的孩子ID数组
 

#include <iostream>
#include <map>
#include <vector>
 
using namespace std;
 
map<int,vector<int> > adjlist;
int levelleaves[]={};
 
void dfs(int node,int level){
    if(adjlist[node].empty()){
        levelleaves[level]++;
        return;
    }
    vector<int>:: iterator itea = adjlist[node].begin();
    for(;itea!=adjlist[node].end();itea++){
        dfs(*itea,level+);
    }
}
 
int main()
{
    int n,m;
    cin>>n>>m;
    int leaves = n-m;
    for(int i=;i<m;i++){
        int id1,k, id2;
        cin>>id1>>k;
        for(int j=;j<k;j++){
            cin>>id2;
            adjlist[id1].push_back(id2);
        }
    }
    dfs(,);
    int a=levelleaves[];
    cout<<levelleaves[];
    for(int i=;a<leaves;i++){
        cout<<" "<<levelleaves[i];
        a=a+levelleaves[i];
    }
    return ;
}