PAT Advanced 1004 Counting Leaves

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, the number of nodes in a tree, and M (<), 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

数所有叶子节点的个数。先根据数据建树，然后dfs确定每个节点的深度并判断是否为叶节点。我的程序里最后又进行了一次bfs统计各个深度下的叶节点个数，其实可以直接在dfs过程中完成。

#include <iostream>
#include <string>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
using namespace std;
struct Node
{
    int depth;
    vector<Node*> children;
    Node(int d):depth(d){};
};
map<int,Node*> nodes;
int N,M,K;

void bfs(Node* ptr,int maxdepth)
{
    queue<Node*> q;
    q.push(ptr);
    int ans[maxdepth+1];
    memset(ans,0,sizeof ans);
    int cnt=0;
    while(!q.empty())
    {
        Node* tmp=q.front();
        q.pop();

        int sz=tmp->children.size();
        if(sz==0)
        {
            cnt++;
            ans[tmp->depth]++;
        }
        for(int i=0;i<sz;i++)
            q.push(tmp->children[i]);
    }
    cout<<ans[0];
    for(int i=1;i<=maxdepth;i++)
        cout<<‘ ‘<<ans[i];
    cout<<endl;
}
int maxdepth=0;
void dfs(Node* ptr,int depth)
{
    if(ptr==NULL)return;
    maxdepth=max(depth,maxdepth);
    ptr->depth=depth;
    for(int i=0;i<ptr->children.size();i++)
    {
        dfs(ptr->children[i],depth+1);
    }
}
int main()
{
    cin>>N>>M;
    if(N==0)return 0;
    if(M==0){cout<<N<<endl;return 0;}
    int tid,tid2;

    for(int i=0;i<M;i++)
    {
        cin>>tid>>K;
        if(nodes.find(tid)==nodes.end())
        {
            nodes[tid]=new Node(0);
        }
        for(int j=0;j<K;j++)
        {
            cin>>tid2;
            if(nodes.find(tid2)==nodes.end())
            {
                nodes[tid2]=new Node(0);

            }
            nodes[tid]->children.push_back(nodes[tid2]);
        }
    }
    dfs(nodes[1],0);
    bfs(nodes[1],maxdepth);
    return 0;
}

原文地址：https://www.cnblogs.com/zest3k/p/11443995.html