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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
IDis a two-digit number representing a given non-leaf node,Kis the number of its children, followed by a sequence of two-digitID's of its children. For the sake of simplicity, let us fix the root ID to be01.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
01is the root and02is its only child. Hence on the root01level, there is0leaf node; and on the next level, there is1leaf node. Then we should output0 1in a line.
这个作为一个30的题，感觉也是很简单的。题目大意就是给出每个非叶子节点的所有孩子，然后求出该树的每一层的叶子节点。已知根节点为1。我处理的思路为先存下每个节点的孩子。然后遍历的时候，寻找出每一层的节点，如果该节点没有孩子了，就说明为叶子节点。这样循环遍历就能求出所有的节点。
#include<bits/stdc++.h>
using namespace std;const int maxn = 106;//建一个树 求每一层的叶子节点个数 01为根节点map<int,vector<int> >node;map<int,vector<int> >vc;//表示前几层 每层的节点
int ans[maxn];
int main()
{int N,M;cin >> N >> M;for(int i = 0;i<M;i++){int rt,k;cin >> rt >> k;for(int j = 0;j<k;j++){int x;cin >> x;node[rt].push_back(x);}}vc[1].push_back(1);//表示根节点只有1 也是第一层int lim = 0;for(int i = 2;i<100;i++){for(int j = 0;j<vc[i-1].size();j++)//拿出上一层节点{int nw = vc[i-1][j];for(int k = 0;k<node[nw].size();k++){int ci = node[nw][k];vc[i].push_back(ci);if(node[ci].size()==0)ans[i]++;//当层存在空姐点}}if(vc[i].size()==ans[i]){lim = i;break;}}if(N==0);else if(N==1){cout << 1;}else {cout << 0;for(int i = 2;i<=lim;i++){cout<<" " << ans[i];}}return 0;
}
)
