PAT-1004 Counting Leaves

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

算法說明：

算法要求求出每一層沒有孩子節點的個數並依次輸出，本文采用vector存儲結點，固然你能夠本身定義結構體。而後用遞歸計算每一層的沒有孩子結點的個數，這時就必須有個數組book[depth]來記錄啦，下標表明層數，對應的值表示這一層無孩子結點個數。遞歸中止的條件爲 :vector[index].size()==0 說明當前結點沒有孩子結點，使book[depth]++，下面貼出代碼。

樹結構與vector存儲

// 1004 Counting Leaves.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <vector>
using namespace std;

vector<int> vec[100];
int maxDepth=-1;
int book[100]={0};
/**
4 2
01 2 02 03
02 1 04

**/
void dfs(int index,int depth){
    if(vec[index].size()==0){
        book[depth]++;
        if(depth>maxDepth)
            maxDepth=depth;
        return;
    }
    for(int i=0;i<vec[index].size();i++)
        dfs(vec[index].at(i),depth+1);
}
int main(int argc, char* argv[])
{
    int N,M,node,k,leaf;
    cin >>N>>M;
    for(int i=0;i<M;i++){
        cin >>node>>k;
        for(int j=0;j<k;j++){
            cin >> leaf;
            vec[node].push_back(leaf);
        }
    }
    dfs(1,0);
    cout<<book[0];
    for(int j=1;j<=maxDepth;j++){
        cout<<" "<<book[j];
    } 
    return 0;
}

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