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<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.
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
题目读起来比较拗口，N节点的数量，M非叶子节点的数量，后面会有M行输入，每行输入内容是，“父节点 子节点数量，子节点序列”
#include <iostream>
#include <list>
using namespace std;
int main(){
//节点数n
int nodeCount;
//非叶子节点数m
int nonLeafCount;
//输入n m
cin >> nodeCount >> nonLeafCount;
//邻接表（index为父节点，list位子节点列表，所以邻接表的size设为最大值100就够了）
list<int> *adj = new list<int>[100];
//输入m行，初始化邻接表
for (int i = 0; i < nonLeafCount; ++i) {
int parent;
int k;
cin >> parent >> k;
for (int j = 0; j < k; ++j) {
int child;
cin >> child;
adj[parent].push_back(child);
}
}
list<int> nodeList;
nodeList.push_back(1);
list<int> resultList;
//逐层处理
while(true){
if (nodeList.empty()){
//empty意味着到达了最深处，不需要继续了
break;
}
int leafCount = 0;
//下一层所有节点的集合
list<int> nextLevelNodeList;
//循环单层所有节点
while(true) {
if (nodeList.empty()){
//这里的empty意味着这一层节点已经全部都判断完毕了
break;
}
int node = nodeList.front();
nodeList.pop_front();
//判断是否有子节点，没有的话leafCount++，有的话，将child加到下一层的节点集合中，下一次遍历
if(!adj[node].empty()){
list<int>::iterator it;
for(it= adj[node].begin(); it != adj[node].end(); it++){
nextLevelNodeList.push_back(*it);
}
} else {
leafCount++;
}
}
nodeList = nextLevelNodeList;
resultList.push_back(leafCount);
}
if(!resultList.empty()){
auto it = resultList.begin();
cout << *it;
it++;
for(; it != resultList.end(); it++){
cout << " " << *it;
}
}
}