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
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
题目大意：
输出树每一行的非叶子节点数
题目解析：
拿到这道题，想到可以用并查集做，开辟三个数组
- bool nonleaf[MAXN]//记录节点是否为叶子节点,初值全为false
- int parent[MAXN]//记录每个节点的父节点序号
- int count[MAXN]//保存每一行的非叶子节点数目,初值为0
每一行读入的数据，开头的ID为非叶子节点，故nonleaf[ID]设为true,后面的节点全为ID的子节点，parent[序号]=ID；
遍历节点数组，递归获取所有nonleaf值为false所在的行号line，对应的count[line]++，同时记录最大行号maxline；
然后输出即可。
具体代码：
#include
using namespace std;
#define MAXN 110
bool nonleaf[MAXN]={false};//noleaf[节点序号]==true表示为非叶子节点
int parent[MAXN];//记录每个节点的父节点序号
int count[MAXN];//保存每一行的非叶子节点数目
int getline(int n){//计算节点所在行号
if(parent[n]==0)
return 1;
return getline(parent[n])+1;
}
int main() {
int n,m;//n(0,100)为节点总数
parent[1]=0;//节点1为根节点
cin>>n>>m;
while(m--){
int pnode,num;//pnode为父节点，num为子节点个数
cin>>pnode>>num;
nonleaf[pnode]=true;//有子节点 ，所以为非叶子节点
while(num--){
int cnode;
cin>>cnode;
parent[cnode]=pnode;//pnode是cnode的父节点
}
}
fill(count,count+MAXN,0);//初始化
int maxline=1;//记录最大行数
for(int i=1;i<=n;i++){
if(nonleaf[i]==false){//如果i是叶子节点，则计算它属于哪一行
int line=getline(i);
count[line]++;//对应行的叶子结点数目加一
if(line>maxline)//若所在行数大于最大行，更新最大行
maxline=line;
}
}
//按要求输出
for(int i=1;i<=maxline;i++){
cout<
发现此题可直接用vector数组+dfs做，正向思维，代码如下：
#include
#include
#define MAXN 110
using namespace std;
vector v[MAXN];
int count[MAXN];
int maxdepth=-1;
void dfs(int index,int depth){
if(v[index].size()==0)//若是叶子节点
{
count[depth]++;
if(depth>maxdepth)
maxdepth=depth;
return;
}
for(int i=0;i>n>>m;
while(m--){
int ID,num;
cin>>ID>>num;
while(num--)
{
int k;
cin>>k;
v[ID].push_back(k);
}
}
fill(count,count+MAXN,0);
dfs(1,0);
for(int i=0;i<=maxdepth;i++)
{
cout<
