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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.
题目大意：给出一个树形结构（家谱），输出每一层的叶子结点的数量。分析：可以使用 bfs，dfs，层序遍历该树，使用 dfs 遍历，用二维数组存储每一个有孩子结点的结点以及他们的孩子，用 leaveNodeCount[depth] 记录每一层的叶结点的数量
#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v[100];
int leaveNodeCount[100];
int maxdepth = -1;
void dfs(int index,int depth){
if(v[index].size() == 0){
leaveNodeCount[depth]++;
maxdepth = max(depth, maxdepth);
return ;
}
for(int i = 0; i < v[index].size(); i++)
dfs(v[index][i],depth+1);
}
int main()
{
int n,m;
scanf(“%d %d”,&n,&m);
for(int i = 0; i < m; i++){
int id,k,child;
scanf(“%d %d”,&id,&k);
for(int j = 0; j < k; j++){
scanf(“%d”,&child);
v[id].push_back(child);
}
}
dfs(1,1);
printf(“%d”,leaveNodeCount[1]);
for(int i = 2; i <= maxdepth; i++){
printf(” %d”,leaveNodeCount[i]);
}
return 0;
}