Problem #
Source: PAT 1004{target="_blank"}
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<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 02Sample Output: #
0 1Solution #
Code #
Github (C++){button.button–outline-info.button–rounded}{target="_blank"}
#include <string>
#include <iostream>
#include <vector>
#include <map>
#include <math.h>
#define max_size 101
using namespace std;
int n, m;
map<int, int> leaf;
struct node
{
int deepth;
vector<int> sons;
node()
{
deepth = 0;
}
};
node tree[max_size];
bool isRoot[max_size];
void dfs(int root)
{
if (leaf.find(tree[root].deepth) == leaf.end())
{
leaf[tree[root].deepth] = 0;
}
if (tree[root].sons.size() == 0)
{
leaf[tree[root].deepth]++;
return;
}
for (int i = 0; i < tree[root].sons.size(); i++)
{
int son = tree[root].sons[i];
tree[son].deepth = tree[root].deepth + 1;
dfs(son);
}
}
int main()
{
cin >> n >> m;
fill(isRoot + 1, isRoot + n + 1, true);
for (int i = 0; i < m; i++)
{
int a, b, c;
string sa, sc;
cin >> sa >> b;
a = (sa[0] - '0') * 10 + sa[1] - '0';
for (int j = 0; j < b; j++)
{
cin >> sc;
c = (sc[0] - '0') * 10 + sc[1] - '0';
tree[a].sons.push_back(c);
isRoot[c] = false;
}
}
int root;
for (root = 1; root <= n; root++)
{
if (isRoot[root])
break;
}
dfs(root);
for (map<int, int>::iterator it = leaf.begin(); it != leaf.end(); it++)
{
if (it != leaf.begin())
cout << " ";
cout << it->second;
}
return 0;
}
