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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:
2 101 1 02Sample Output:
0 1本题思路：用map来映射所有父节点和子节点之间的关系，并且此题已经确定根节点为01，因此可以逐层遍历寻找
AC代码：
#include<iostream>#include<algorithm>#include<unordered_map>#include<vector>using namespace std;int n,m;unordered_map<int,vector<int>> tree;int main(){ scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { int t,k,p; scanf("%d%d",&t,&k); vector<int> ans={ }; for(int j=0;j<k;j++) { scanf("%d",&p); ans.push_back(p); } tree[t]=ans; } vector<int> res=tree[1]; if(n==1) printf("1"); else { int cnt=1; printf("0"); while(cnt<n) { vector<int> transition={ }; int count=0; for(int i=0;i<res.size();i++) { if(tree.count(res[i])) { for(int j=0;j<tree[res[i]].size();j++) transition.push_back(tree[res[i]][j]); } else count++; } cnt+=res.size(); res.assign(transition.begin(),transition.end()); printf(" %d",count); } } return 0;}转载地址：http://jzbwz.baihongyu.com/