PAT 1004 Counting Leaves(分层+统计每层叶子节点数）

PAT1004

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

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：给一个多叉树，统计每层的叶子节点的数目。

思路：用邻接表将多叉树存图，第一个dfs将图进行分层，第二个dfs统计叶子节点的个数。

#include
#include
#include
#include
#define memset(a,v) memset(a,v,sizeof(a))
using namespace std;
vectorvec[1100];
int level[1100];
int cnt[1100];
int maxLevel;
void init() {
    maxLevel=0;
    for(int i=0;i<=200;i++) vec[i].clear();
    memset(level,0);
    memset(cnt,0);
}
void getLevel(int u) {
    for(int i=0;i