PAT（甲级）1004.Counting Leaves（30）
PAT（甲级）1004.Counting Leaves（30）
PAT 1004.Counting Leaves（30） A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
输入格式：
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.
输出格式：
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 1leaf node. Then we should output 0 1 in a line.
输入样例：
2 1
01 1 02输出样例：
0 1题目分析：输出多叉树中每层中叶子节点的数目。可以使用dfs或bfs。由于题目中提示了节点编号，可以使用静态树结构，使用邻接表的方式存储每个节点的孩子节点的编号。其实也是一种hash的思想，hash[parentIndex] = {}，只不过其对应的是一个集合。
使用bfs时，当前队列中的元素即为当前层上的所有节点，只需要枚举每个节点是否有孩子，如果有就将其所有孩子都继入队列，如果没有在则当前层的叶子节点个数+1。
AC代码（bfs）：
#include <queue>
#include <vector>
using namespace std;
vector<int> tree[100];//邻接表存储树
int floor[100];//每层叶子节点的个数
int index=0;//根节点在第0层
void bfs(int root){
queue<int> q;
q.push()；
while(!q.empty()){
int size = q.size();
//注意不能写成for(int i=0; i<q.size(); ++i)这样q.size()是会改变的
//因为每次枚举一个队首元素 都会出队 所以q.size() -= 1;
for(int i=0; i<size; ++i){
int front = q.front();q.pop();
if(!tree[front].size()){ //没有孩子节点 说明是叶子节点
floor[index] ++;
}
else{
//不是叶子节点 将其所有的孩子节点入队
for(int j=0; j<tree[front].size(); ++j){
q.push(tree[front][j]);
}
}
}
index ++;//下一层
}
}
int main(){
int n, m, k, p, c;
scanf("%d%d", &n, &m);
for(int i=0; i<m; ++i){
scanf("%d %d", &p, &k);
while(k --){
scanf("%d", &c);
tree[p].push_back(c);
}
}
bfs(1);
for(int i=0; i<index; ++i){
printf("%d", floor[i]);
if(i!=index-1)printf(" ");
}
return 0;
}AC代码(dfs)：
vector<int> tree[100];
int floor[100];
int maxDepth=-1;//这个变量用于确定该树有几层
//dfs需要传入的参数 当前节点的索引 当前节点所在的深度
void dfs(int p, int depth){
//递归出口
//没有孩子节点 说明找到一个叶子节点
if(!tree[p].size()){
floor[depth] ++;//当前层叶子节点个数+1
//更新最大的深度
maxDepth = max(depth, maxDepth);
return ;
}
else{
//不是叶子节点，枚举其所有孩子节点是否为叶子节点
for(int i=0; i<tree[p].size(); ++i)
dfs(tree[p][i], depth+1); //孩子节点的深度比当前节点的深度大1
}
}
int main(){
int n, m, k, p, c;
scanf("%d%d", &n, &m);
for(int i=0; i<m; ++i){
scanf("%d %d", &p, &k);
while(k --){
scanf("%d", &c);
tree[p].push_back(c);
}
}
dfs(1, 0);//根节点的索引为1 根节点所在层为0
for(int i=0; i<=maxDepth; ++i){
printf("%d", floor[i]);
if(i!=maxDepth)printf(" ");
}
return 0;
}腾讯云开发者
