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<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 01is 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输入：总节点数n，非叶节点数 m
m行：给定的非叶节点 ID，非叶节点孩子节点的个数 k, 每个孩子的 ID[1],.........ID[k]
输出：每一层的叶节点数
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v[100];
int book[100], maxdepth = -1;
void dfs(int index, int depth) {
//出口
if(v[index].size() == 0) {//若当前节点为空
book[depth]++; //上一层空节点数++
maxdepth = max(maxdepth, depth);
return ;
}
for(int i = 0; i < v[index].size(); i++)
dfs(v[index][i], depth + 1);//dfs每个index的孩子节点
}
int main() {
int n, m, k, node, c;
scanf("%d %d", &n, &m);
for(int i = 0; i < m; i++) {
scanf("%d %d",&node, &k);//非叶结点编号 以及其孩子个数
for(int j = 0; j < k; j++) { //输入每个孩子的编号
scanf("%d", &c);
v[node].push_back(c);//将c拷贝到vector最后一个元素的后面
}
}
dfs(1, 0);//从第一个节点，第0层开始
printf("%d", book[0]);
for(int i = 1; i <= maxdepth; i++)
printf(" %d", book[i]);
return 0;
}