标签:Counting families level int Leaves child 1004 root ID
1004 Counting Leaves
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
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
struct family
{
int level;
vector<int> child;
}families[110];
int total;
int non_leaf;
int ID;
int id;
int num_of_child;
int no_child[110];
int maxlevel = 0;
void level(int root)
{
queue<family> Q;
families[root].level = 1;
Q.push(families[root]);
while (!Q.empty())
{
family p = Q.front();
Q.pop();
for (int i = 0; i < p.child.size(); i++)
{
int chi = p.child[i];
families[chi].level = p.level + 1;
Q.push(families[chi]);
}
}
}
int main()
{
scanf_s("%d %d", &total, &non_leaf);
for (int i = 0; i < non_leaf; i++)
{
scanf_s("%d %d", &ID,&num_of_child);
for (int j = 0; j < num_of_child; j++)
{
scanf_s("%d", &id);
families[ID].child.push_back(id);
}
}
level(1);
for (int i = 1; i <= total; i++)
{
if (families[i].child.size() == 0)
{
no_child[families[i].level]++;
}
if (families[i].level > maxlevel)
{
maxlevel = families[i].level;
}
}
for (int i = 1; i <= maxlevel; i++)
{
printf("%d", no_child[i]);
if (i < maxlevel)
{
printf(" ");
}
else
{
printf("\n");
}
}
}
标签:Counting,families,level,int,Leaves,child,1004,root,ID 来源: https://blog.csdn.net/weixin_44230791/article/details/123231489
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