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.

Input

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.

Output

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

2 101 1 02

Sample Output

0 1

对于我这种小白来说，每次都被题目的意思弄崩溃。所以……

这道题的意思就是这是一个家族树，给的N,M是树上所有的结点数，和不是叶子结点的结点数（non-leaf ：不能翻译为没有叶子的结点 别问我是怎么知道的）

这里 ID 就是这个结点的代号 K就是他孩子的个数 后面就是这k个孩子的代号

输出要求：求出每一层鳏夫寡妇的个数（孩子结点的喽）

代码还是仿照网上大牛的方法写的：

代码核心就是深搜一下，深搜的层数就是当前结点的层数，定义一个全局数组用来保存每一层寡妇的个数……

#include<cstdio>#include<map>#include<vector> #include<iostream>using namespace std;int levels[101];map<int,vector<int> > tree;//map ：关联容器（注释给和我一样的小白瞅~不谈） void dfs(int id,int level)//深搜  {if(tree[id].empty())levels[level]++;else{vector<int>::iterator it =tree[id].begin();//迭代器 for(;it!=tree[id].end();it++)dfs(*it,level+1);}}int main(){int n,m,i,j,id1,id2,k,count;cin>>n>>m;for(i=0;i<m;i++){cin>>id1>>k;for(j=0;j<k;j++){cin>>id2;tree[id1].push_back(id2);//关联容器赋值的最懒法子~ }}dfs(1,0);cout<<levels[0];count=levels[0];for(i=1;count!=(n-m);i++){cout<<" "<<levels[i];count+=levels[i];}return 0;}