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; }