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
題目大意：
輸入： 第一行有n和m。n代表一共有多少個結點，m代表一共有多少個非葉結點。接下來一共m行，格式爲：
ID K ID[1] ID[2] … ID[K]
ID表示非葉結點的編號，K代表這個非葉結點一共有多少個孩子，接下來的ID[1…k]爲這k個孩子的編號。
輸出： 輸出每一層沒有孩子結點的結點個數。
思路：利用一個結構體記錄每一個結點的父結點，所在層數，有無孩子。在輸入的過程中會記錄每一個孩子的父親結點以及這個結點是否有孩子。層數可以在遍歷的時候記錄（已經知道根結點是第一層）。
參考文章
#include<iostream>
using namespace std;
struct node {
int level=0;
int father=0;
bool nochild=1;
};
node a[105];
int l[105];
int main() {
int n,m,k,c,id,maxlevel=1;
scanf("%d %d",&n,&m);
for(int i=1; i<=m; i++) {
scanf("%d %d",&id,&k);
a[id].nochild=0;
for(int j=1; j<=k; j++) {
scanf("%d",&c);
a[c].father=id;
}
}
a[1].level=1;
//接下來遍歷，記錄每一個結點所在的層數以及最大層數（後面輸出會用到最大層數）
for(int i=1; i<=n; i++) {
for(int j=1; j<=n; j++) {
if(a[j].father==i) {
a[j].level=a[i].level+1;
if(a[j].level>maxlevel) maxlevel=a[j].level;
}
}
}
//遍歷計算每層沒有孩子結點的結點個數
for(int i=1; i<=n; i++) {
if(a[i].nochild==1) {
l[a[i].level]++;
}
}
for(int i=1; i<=maxlevel; i++) {
if(i==1)
printf("%d",l[i]);
else printf(" %d",l[i]);
}
}