pat甲级 1004 countleaves （二维向量容器存储树结构 + dfs + bfs）

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：

        可以通过dfs 和 bfs两种方式实现：

        dfs 思路：将一条路走通， 当走不通时再回溯
        bfs 思路：每次走一步，将所有一步可以到达的路径先后放入队列中，当队列不为空时，将队列的第一个元素出队，

        当找到一个结点没有子结点时，叶子个数+1；

        用二维向量容器存储树：a[i][j] = k;//表示 id 为 i的结点的第j个孩子结点的 id为k

  dfs代码实现：

       

#include<iostream>
#include<vector>
//本题思路就是判断 每层的叶子结点（不存在其它的子结点）有多少个 
using namespace std;
vector<int>a[100]; //定义二维向量数组存储树（图）结构 
int n, m, hmax = -1, h, level[100]; // 也可不做处理 ，hmax = -1 题目中告诉了 n = 1时不做处理 
									//level[i]表示存储高度为i的叶子结点个数 

int pid, k, cid; //表示表此时结点的id 子结点的个数，孩子结点的Id 
void dfs(int start, int h){ //dfs遍历  （start,h） 
	hmax = max(hmax, h);
	if(a[start].size() == 0) { //当size == 0表示结点没有连接任何一个结点 表示不存在下一个结点 此时此高度叶子结点数++ 
		level[h]++;      //h为此时的高度， 
		return;
	}
	for(int i = 0; i < a[start].size(); i++){
		dfs(a[start][i], h + 1);//保留此时的h高度 ，回溯时高度依然是此时的高度 
	}
}
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 0; i < m; i++){
		scanf("%d%d", &pid, &k);
		for(int j = 0; j < k; j++){
			scanf("%d", &cid);
			a[pid].push_back(cid); 
		}
	}
	dfs(1, h);
	for(int i = 0; i <= hmax; i++){
		if(i != 0)
			printf(" ");
		printf("%d", level[i]);
	}
	return 0;
} 

bfs代码实现：

#include<iostream>
#include<queue>
#include<vector>
/**
dfs 思路：将一条路走通， 当走不通时再回溯
bfs 思路：每次走一步，将所有一步可以到达的路径先后放入队列中，当队列不为空时，将队列的第一个元素出队，
//判断是否有一步可以到达的点，将其放入队列中. 
**/
using namespace std;
struct Node{
	int to;
	int h;
};
int n, m;
int pid, k, cid, hmax, level[100];
vector<int> a[100];      
queue<Node> q;        //将对应结点的编号加入进队列中 
void bfs(int start){  //当 
	q.push({start, 0});
	while(!q.empty()){
		Node node = q.front();
		q.pop();	
		if(a[node.to].size()==0){
			level[node.h]++;  // 
			continue;
		} 
		hmax = node.h + 1;    //当结点不为叶子节点时，此叶子结点的高度为此时结点的高度 + 1  
		for(int i = 0; i < a[node.to].size(); i++){
			q.push({a[node.to][i], hmax});
		}
	}
	 
}
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 0; i < m; i++){
		scanf("%d%d", &pid, &k);
		for(int j = 0; j < k; j++){
			scanf("%d", &cid);
			a[pid].push_back(cid);
		}
	}
	bfs(1);
	for(int i = 0; i <= hmax; i++){
		if(i != 0)
			printf(" ");
		printf("%d", level[i]);
	}
	return 0;
}