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  • 1004 Counting Leaves (30分) DFS

    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


    题意:
      给一棵树,求这棵树中每一层中,没有子节点的节点个数。第一行输入两个数n,m;分别表示这棵树节点和叶子节点个数,第二行依次输入节点编号和这个节点子节点个数

    题解:
      用vector数组保存每个节点的儿子节点,然后用DFS依次遍历每个深度的每个节点,数组记录每层子节点数为0的节点即可

    #include<iostream>
    #include<cstdio>
    #include<vector>
    #include<cstring>
    #define MAX 1000000
    using namespace std;
    int num[1005],vis[10005];
    vector<int>v[1005];
    int cnt=0;
    void dfs(int root,int dep)
    {
        vis[root]=1;
        if(v[root].size()==0)
        {
            num[dep]++;
            cnt=cnt<dep?dep:cnt;//记录最大深度
        }
        else
        {
            for(int i=0;i<v[root].size();i++)//遍历儿子节点
            {
                if(vis[v[root][i]]==0)
                    dfs(v[root][i],dep+1);
            }
        }
    }
    int main()
    {
        int n,m,id,k;
        cin>>n>>m;
        for(int i=0;i<m;i++)
        {
            cin>>id>>k;
            for(int i=0;i<k;i++)
            {
                int chi;//儿子节点
                cin>>chi;
                v[id].push_back(chi);
            }
        }
        dfs(1,0);
        for(int i=0;i<=cnt;i++)
        {
            if(i==0)
                cout<<num[i];
            else
                cout<<' '<<num[i];
        }
        cout<<endl;
        return 0;
    }

      
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  • 原文地址:https://www.cnblogs.com/-citywall123/p/12095289.html
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