1004 Counting Leaves （30 point(s)）

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

題目大意:

給定對應的樹的結構.然後給出每一層的葉子節點的數量.

解決:

1.首先是輸入的轉化,要將01 ...02 等等數字轉化成正常的數字,解決方案就是使用sstream  通過string轉化為int

2.其次是樹的儲存和構造,這裡用了vector陣列進行儲存,因為知道範圍,所以直接建立是很方便的

3.最後是樹的層序遍歷,使用queue進行層序遍歷每一層遍歷完之後,直接列印對應的葉子節點的數量

4.注意最後的列印格式

#include<iostream>
#include<vector>
#include<sstream>
#include<string>
#include<queue>
using namespace std;


int strToint(const string &str){
    istringstream i(str);
    int out;
    if(i>>out){
        return out;
    }
    return 0;
}

int main(){
    int const MAX = 110;
    bool vis[MAX]{false};
    int father[MAX];
    vector<int>tree[MAX];
    queue<int>finder;

    fill(father,father+MAX,1);
    fill(vis,vis+MAX,0);

    int N=0;
    int M=0;

    cin>>N>>M;

    if(N==0){
        return 0;
    }

    for(int i=0;i<M;i++){
        string temp;
        int num;
        cin>>temp>>num;
        int id = strToint(temp);
        vis[id] =true;
        for(int j=0;j<num;j++){
            string temp_num;
            cin>>temp_num;
            int temp_id = strToint(temp_num);
            vis[temp_id] = true;
            tree[id].push_back(temp_id);
            father[temp_id] = id;
        }
    }
    finder.push(1);
    int Level =1;
    int oldlevel =1;
    int Count=0;
    bool first=true;
    while(!finder.empty()){
        int id = finder.front();
        finder.pop();
        int lenth = tree[id].size();
        if(lenth==0){
            Count++;
        }
        for(int i=0;i<lenth;i++){
            finder.push(tree[id][i]);
            Level = tree[id][i];
        }

        if(id==oldlevel){
            oldlevel = Level;
            if(first){
                cout<<Count;
                first = false;
            }else{
                cout<<" "<<Count;
            }
            Count=0;
        }

    }

    return 0;

}

一開始理解錯了,以為是要找到葉子節點並且最後列印從root到該葉子節點的路徑...審題很總要

生僻單詞:

hierarchy 

n. 層級；等級制度

pedigree

• n. 血統；家譜
• adj. 純種的

specification

n. 規格；說明書；詳述

sake

• n. 目的；利益；理由；日本米酒
• n. (Sake)人名；(羅)薩克；(日)酒(姓)

seniority

n. 長輩；老資格；前任者的特權