1004. Counting Leaves

pat甲级

思路：分类统计。题目是统计树的每一层叶子的个数。遍历树，如果当前结点是叶子，就加入到相应的统计中。用数组保存每层叶子数。注意，数据中包含了00这个结点。

训练目的：树的遍历

题目描述

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

参考代码

#include <iostream>
#include <vector>
using namespace std;
#include <string.h>

int n;
vector<int> tree[100];
int leaves[100];
int maxLevel=0;
void traersal(int root,int level)
{
  if (maxLevel<level) maxLevel=level;
  if (tree[root].size()==0)
    leaves[level]++;
  else
    for(vector<int>::iterator it=tree[root].begin();it!=tree[root].end();it++)
      traersal(*it,level+1);
}
int main()
{
  int m;
  while(cin>>n>>m)
  {
    int i;
  for(i=0;i<100;i++)
    tree[i].clear();
    for(i=0;i<m;i++)
    {
    char num[3];
    cin>>num;;
    int id=atoi(num);
    int k;
    cin>>k;
    for(int j=0;j<k;j++)
    {
      cin>>num;
      tree[id].push_back(atoi(num));
    }
    }
    memset(leaves,0,sizeof(int)*100);
  maxLevel=0;
  traersal(1,0);
  cout<<leaves[0];
  for(i=1;i<=maxLevel;i++)
    cout<<" "<<leaves[i];
  cout<<endl;
  }
  return 0;
}