1004 Counting Leaves (30分)（一个字短！）

1004 Counting Leaves (30分)

题目

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

题意

用邻接表输入一颗树，输出每层的叶子节点的数目

分析

层序遍历或者深度优先遍历

代码

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
int n, m, id, k, a;
vector<int> tree[110], v;
int main() {
 	scanf("%d %d", &n, &m);
 	for (int i = 0; i < m; ++i) {
  		scanf("%d %d", &id, &k);
  		for (int j = 0; j < k; ++j) {
   			scanf("%d", &a);
   			tree[id].push_back(a);
  		}
 	}
 	int cnt;
 	queue<int> q;
 	q.push(1);
 	while (!q.empty()) {
  		queue<int> p;
  		cnt = 0;
  		while (!q.empty()) {
   			int f = q.front();
   			q.pop();
   			if (tree[f].size() == 0)++cnt;
   			for (int i = 0; i < tree[f].size(); ++i)p.push(tree[f][i]);
  		}
  		q = p;
  		v.push_back(cnt);
 	}
 	for (int i = 0; i < v.size(); ++i)
  		printf("%d%s", v[i], i == v.size() - 1 ? "" : " ");
 	return 0;
}