PAT-Advance 1004. Counting Leaves (30)

https://www.patest.cn/contests/pat-a-practise/1004

1004. Counting Leaves (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input

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.

Output

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

解析：问题实质其实是得到此树每一层的叶子节点数。

通过广度优先算法遍历整个树，我们便能得到结果。

#include <iostream>
#include <map>
#include <vector>

using namespace std;

void getLeafs(int N, int M, map<int, vector<int>> &mapTree) {
	if (mapTree.find(1) == mapTree.end()) {  //判断根节点是否有子节点
		cout << 1 << endl;
		return;
	}
	vector<int> rst; //用于保存每一层的叶子节点数目
	rst.push_back(0);
  	cout << 0;
	vector<int> tmpvec(mapTree[1]); //用于保存每一层的节点
	while (!tmpvec.empty())
	{

		int size = tmpvec.size();
		int levelCount = 0;
		for (int i = 0; i < size; ++i) {
			int head = tmpvec[0];
			if (mapTree.find(head) != mapTree.end()) {  //有子节点，则将子节点插入此队列
				//levelCount++;
				for (auto itor = mapTree[head].begin(); itor != mapTree[head].end(); ++itor) {
					tmpvec.push_back(*itor);
				}
				tmpvec.erase(tmpvec.begin()); //此节点处理完之后，将其从队首移除
			}
			else {
				tmpvec.erase(tmpvec.begin()); //此节点处理完之后，将其从队首移除
				levelCount++; //无子节点，本层计数加一
			}
		}
		rst.push_back(levelCount);
	}
	for (auto itor = rst.begin() + 1; itor != rst.end(); ++itor) {
		cout << " " << *itor;
	}
}

int main() {
	//freopen("PAT-A1004.txt", "r", stdin);
	int N, M;
	cin >> N >> M;
	map<int, vector<int>> mapTree;
	int ID, K;
	vector<int> chdVec;
	for (int i = 0; i < M; ++i) {
		
		cin >> ID >> K;
		for (int j = 0; j < K; ++j) {
			int tmp;
			cin >> tmp;
			chdVec.push_back(tmp);
		}
		mapTree[ID] = chdVec;
		chdVec.clear();
	}

	getLeafs(N, M, mapTree);

	//system("pause");
	return 0;
}