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PAT (Advanced Level) 1004 Counting Leaves (30 分)

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 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

code

#include<cstdio>
#include<cstring>
struct Node
{
	int father;
	int level;
	bool NoChild;
};
Node v[100]; //用以記錄樹的資訊
char level[100] = "\0"; //用以記錄樹每層葉子節點數
int main()
{
	int N, M, c, ID, child, MAXLevel = 1;
	scanf_s("%d%d", &N, &
M); //初始化 for (int i = 0; i<100; ++i) { v[i].father = 0; v[i].level = 0; v[i].NoChild = 1; } //獲取輸入第2行到第M行的資訊 for (int i = 0; i<M; ++i) { scanf_s("%d%d", &ID, &c); v[ID].NoChild = 0; for (int j = 0; j<c; ++j) { scanf_s("%d", &child); v[child].father = ID; } } v[1].level = 1; //根節點層數初始化為1 for (int i = 1; i <= N; ++i) for (int j = 1; j <= N; ++j) { //如果一個節點的父節點是i if (v[j].father == i) { v[j].level = v[i].level + 1;//那麼他的層數等於父節點層數+1 if (v[j].level>MAXLevel) MAXLevel = v[j].level; //維護一個MAXLevel記錄最大層數 } } for (int i = 1; i <= N; ++i) //統計每一層的葉子節點數 if (v[i].NoChild == 1) level[v[i].level]++; //輸出 for (int i = 1; i<MAXLevel; ++i) printf("%d ", level[i]); printf("%d\n", level[MAXLevel]); return 0; }

思路

演算法來自於這篇部落格https://blog.csdn.net/qq278672818/article/details/54915636只對它的程式碼做了一點小改動和加了註釋,感謝博主的分享。本來是準備建一棵樹然後用廣度優先搜尋去做的,但是在測試點1和3上出現段錯誤,後來考慮到輸入的過程可能是亂序的,比如考慮構建如下一棵樹:
在這裡插入圖片描述輸入為
7 4
01 1 02
02 3 03 04 05
05 1 06
06 1 07
這樣順序輸入就是對的,不會段錯誤。
但是輸入為
7 4
01 1 02
05 1 06
02 3 03 04 05
06 1 07
因為處理05 1 06時找不到05號節點而產生段錯誤
於是我對程式碼做了調整,在找不到父親節點時就新建一棵以這個節點為根的樹,加入trees的list,而且每次建子節點之前遍歷這個list看看是否有根節點ID與子節點ID相同的,如果有就把這兩棵樹連起來,但只對了測試點3,測試點1還是段錯誤,不知道哪裡還有問題。。
這是原始程式碼,錯測試點1和3。

#include <iostream>
#include <map>
#include <string>
#include <queue>

using namespace std;

struct Node
{	
	string id;
	int level;
	vector<Node*> children;
	Node(string id, int level);
	void getChild(string id, int plevel);
};

struct Tree
{
	Node* root;
	Node* find(string id);
	Tree();
	map<int, int> count();
};

Tree::Tree()
{
	root = new Node("01", 0);
}
Node* Tree::find(string id)
{
	queue<Node*> nodeQueue;
	nodeQueue.push(root);
	while (!nodeQueue.empty())
	{
		Node* node = nodeQueue.front();
		nodeQueue.pop();
		if (node->id == id)
			return node;
		else
		{
			for (size_t i = 0; i < node->children.size(); i++)
			{
				nodeQueue.push(node->children[i]);
			}
		}
	}
	return nullptr;
}
map<int, int> Tree::count()
{
	map<int, int> countMap; // <level, num>
	queue<Node*> nodeQueue;
	nodeQueue.push(root);
	while (!nodeQueue.empty())
	{
		Node* node = nodeQueue.front();
		nodeQueue.pop();
		if (!countMap.count(node->level))
			countMap[node->level] = 0;
		if (node->children.size() == 0)
			countMap[node->level]++;
		for (size_t i = 0; i < node->children.size(); i++)
		{
			nodeQueue.push(node->children[i]);
		}
	}
	return countMap;
}

Node::Node(string id, int level)
{
	this->id = id;
	this->level = level;
}
void Node::getChild(string id, int plevel)
{
	Node* childNode = new Node(id, plevel + 1);
	children.push_back(childNode);
}
int main()
{
	int numOfNode;
	for (; cin >> numOfNode; )
	{
		if (!numOfNode) break;
		Tree tree;
		int numOfNonLeaf;
		cin >> numOfNonLeaf;
		for (size_t i = 0; i < numOfNonLeaf; i++)
		{
			string pid;
			int k; // num of children
			cin >> pid >> k;
			Node* pNode = tree.find(pid);
			for (int j = 0; j < k; j++)
			{
				string cid;
				cin >> cid;
				pNode->getChild(cid, pNode->level);
			}
		}
		map<int, int> countMap = tree.count();
		cout << countMap.begin()->second;
		map<int, int>::iterator it = countMap.begin();
		it++;
		for (; it != countMap.end(); it++)
			cout << " " << it->second;
		cout << endl;
	}
	return 0;
}

這是修改版程式碼錯測試點1

#include <iostream>
#include <map>
#include <string>
#include <queue>
#include <list>

using namespace std;

struct Node
{	
	string id;
	int level;
	vector<Node*> children;
	Node(string id, int level);
	void getChild(string id, int plevel);
};

struct Tree
{
	Node* root;
	Node* find(string id);
	Tree();
	Tree(string id);
	map<int, int> count();
	void updateLevel();
};

Tree::Tree()
{
	root = new Node("01", 0);
}
Tree::Tree(string id)
{
	root = new Node(id, 0);
}
Node* Tree::find(string id)
{
	queue<Node*> nodeQueue;
	nodeQueue.push(root);
	while (!nodeQueue.empty())
	{
		Node* node = nodeQueue.front();
		nodeQueue.pop();
		if (node->id == id)
			return node;
		else
		{
			for (size_t i = 0; i < node->children.size(); i++)
			{
				nodeQueue.push(node->children[i]);
			}
		}
	}
	return nullptr;
}
map<int, int> Tree::count()
{
	map<int, int> countMap; // <level, num>
	queue<Node*> nodeQueue;
	nodeQueue.push(root);
	while (!nodeQueue.empty())
	{
		Node* node = nodeQueue.front();
		nodeQueue.pop();
		if (!countMap.count(node->level))
			countMap[node->level] = 0;
		if (node->children.size() == 0)
			countMap[node->level]++;
		for (size_t i = 0; i < node->children.size(); i++)
		{
			nodeQueue.push(node->children[i]);
		}
	}
	return countMap;
}
void Tree::updateLevel()
{
	queue<Node*> nodeQueue;
	nodeQueue.push(root);
	while (!nodeQueue.empty())
	{
		Node* node = nodeQueue.front();
		nodeQueue.pop();
		for (size_t i = 0; i < node->children.size(); i++)
		{
			node->children[i]->level = node->level + 1;
			nodeQueue.push(node->children[i]);
		}
	}
	return;
}
Node::Node(string id, int level)
{
	this->id = id;
	this->level = level;
}
void Node::getChild(string id, int plevel)
{
	Node* childNode = new Node(id, plevel + 1);
	children.push_back(childNode);
}
int main()
{
	int numOfNode;
	for (; cin >> numOfNode; )
	{
		if (!numOfNode) break;
		list<Tree> trees;
		int numOfNonLeaf;
		cin >> numOfNonLeaf;
		for (size_t i = 0; i < numOfNonLeaf; i++)
		{
			string pid;
			int k; // num of children
			cin >> pid >> k;
			Node* pNode = nullptr;
			for (auto it : trees)
			{
				pNode = it.find(pid);
				if (pNode != nullptr) break;
			}
			if (pNode == nullptr)
			{
				Tree tree(pid);
				trees.push_back(tree);
				pNode = tree.root;
				for (int j = 0; j < k; j++)
				{
					string cid;
					cin >> cid;
					Node* cNode = nullptr;
					list<Tree>::iterator it = trees.begin();
					for (; it != trees.end(); it++)
					{
						if (it->root->id == cid)
						{
							cNode = it->root;
							trees.erase(it);
							break;
						}
					}
					if (cNode != nullptr)
					{
						pNode->children.push_back(cNode);
					}
					else
					{
						pNode->getChild(cid, pNode->level);
					}
				}
			}
			else
			{
				for (int j = 0; j < k; j++)
				{
					string cid;
					cin >> cid;
					Node* cNode = nullptr;
					list<Tree>::iterator it = trees.begin();
					for (; it != trees.end(); it++)
					{
						if (it->root->id == cid)
						{
							cNode = it->root;
							trees.erase(it);
							break;
						}
					}
					if (cNode != nullptr)
					{
						pNode->children.push_back(cNode);
					}
					else
					{
						pNode->getChild(cid, pNode->level);
					}
				}
			}
		}
		trees.begin()->updateLevel();
		map<int, int> countMap = trees.begin()->count();
		cout << countMap.begin()->second;
		map<int, int>::iterator it = countMap.begin();
		it++;
		for (; it != countMap.end(); it++)
			cout << " " << it->second;
		cout << endl;
	}
	return 0;
}

以上