PAT 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

思路

输出每个高度叶子节点的个数。

深搜，搜到底后，使本高度的叶子节点数加一。

需要注意的是，我被卡了一组数据，因为我判断的是叶子节点的方法为mp[i].size() == 1。

实际上，当根节点只有一个子节点时，也满足上述条件，因此需要特判一下。

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;

int N, M;
vector<int> mp[110];
int num[110];
int vis[110];
int maxd;

void dfs(int index, int deep){
	if(index != 1 && mp[index].size() == 1){
		num[deep]++;
		maxd = max(deep, maxd);
		return;
	}
		
	for(int i = 0; i < mp[index].size(); i++){
		if(!vis[mp[index][i]]){
			vis[mp[index][i]] = 1;
			dfs(mp[index][i], deep + 1);
		}
	}
}


int main(){
	cin >> N >> M;
	for(int i = 0; i < M; i++){
		int id = 0, k = 0;
		cin >> id >> k;
		int t = 0;
		for(int j = 0; j < k; j++){
			cin >> t;
			mp[t].push_back(id);
			mp[id].push_back(t);
		}
	}
	
	vis[1] = 1;
	dfs(1, 1);
	
	if(mp[1].size() == 0)	cout << "1" << endl;
	
	for(int i = 1; i <= maxd; i++){
		if(i == maxd)
			cout << num[i];
		else
			cout << num[i] << " ";
	}
//	for(int i = 0; i < 100; i++){
//		if(!mp[i].size())	continue;
//		cout << i << " ";
//		for(int j = 0; j < mp[i].size(); j++){
//			cout << mp[i][j] << " ";
//		}
//		cout << endl;
//	}
	return 0;
}