PAT (Advanced Level) Practice 1004

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

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

分析：需要求一棵树每一层的叶子节点个数。用二维vector来储存树，BFS树来求解。用结构体记录节点与对应层数，也方便记录最大层数。BFS中每一次将子节点入队，将节点对应从层数+1。若某一节点无子节点，则将其对应层数叶子节点数+1。最后输出结果即可。

代码： 

#include
#include
#include
#include
#define MAXN 101
using namespace std;
int N, M;
int inQueue[MAXN] = { 0 };
int ans[MAXN] = { 0 };
struct Node {
	int value;
	int level;
};
vector v[MAXN];
int maxLevel = 1;
void bfs() {
	queue q;
	q.push({1, 1});
	inQueue[1] = 1;
	while (q.empty() != 1) {
		Node top = q.front();
		maxLevel = max(maxLevel, top.level);
		q.pop();
		if (v[top.value].size() == 0) {
			ans[top.level]++;
		} else {
			for (int i = 0; i < v[top.value].size(); i++) {
				q.push({ v[top.value][i], top.level + 1});
				inQueue[v[top.value][i]] = 1;
			}
		}
	}
}
int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int id;
		int K;
		cin >> id >> K;
		for (int j = 0; j < K; j++) {
			int temp;
			cin >> temp;
			v[id].push_back(temp);
		}
	}
	bfs();
	for (int i = 1; i <= maxLevel; i++) {
		if (i != maxLevel) {
			cout << ans[i] << ' ';
		} else {
			cout << ans[i] << endl;
		}
	}
	return 0;
}