Source:

PAT A1004 Counting Leaves (30 分)

Description:

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, the number of nodes in a tree, and M (<), 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

Keys:

Code:

 /*
time: 2019-06-28 16:47:33
problem: PAT_A1004#Counting Leaves
AC: 16:22 题目大意:
统计各层的叶子结点个数
输入:
第一行给出,结点数N<100,分支结点数M
接下来M行,结点id,孩子数k,孩子id(1~n,root==1)
多组输入样例 基本思路:
遍历并记录各层叶子结点数即可
*/
#include<cstdio>
#include<vector>
using namespace std;
const int M=;
vector<int> tree[M];
int leaf[M],level=; void Travel(int root, int hight)
{
if(tree[root].size()==)
{
if(hight > level)
level = hight;
leaf[hight]++;
return;
}
for(int i=; i<tree[root].size(); i++)
Travel(tree[root][i],hight+);
} int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE int n,m;
while(~scanf("%d", &n))
{
level=;
for(int i=; i<=n; i++){
tree[i].clear();
leaf[i]=;
}
scanf("%d", &m);
for(int i=; i<m; i++)
{
int id,k,kid;
scanf("%d%d", &id,&k);
for(int j=; j<k; j++)
{
scanf("%d", &kid);
tree[id].push_back(kid);
}
}
Travel(,);
for(int i=; i<=level; i++)
printf("%d%c", leaf[i], i==level?'\n':' '); } return ;
}

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