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
题目大意：
求每一层上的叶子结点个数。
思路：
用bfs
代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#define MaxSize 105
using namespace std;
struct Node
{
int k;
int leaf[MaxSize];
};
int layer;
int leafCount[MaxSize];
Node *nodes[MaxSize];
void bfs()
{
int seq[MaxSize];
int front = 0, end = 0;
int k, l; //当前层和下一层的结点个数
Node *t;
layer = 0;
seq[end] = 1;
end = (end + 1) % MaxSize;
k = 1;
l = 0;
while (front != end)
{
t = nodes[seq[front]];
front = (front + 1) % MaxSize;
if (t->k == 0)
{
leafCount[layer]++;
}
else
{
l += t->k;
for (int i = 0; i < t->k; i++)
{
seq[end] = t->leaf[i];
end = (end + 1) % MaxSize;
}
}
k--;
if (k == 0)
{
layer++;
k = l;
l = 0;
}
}
}
int main()
{
memset(leafCount, 0, sizeof(leafCount));
for (int i = 0; i < MaxSize; i++)
{
nodes[i] = new Node();
nodes[i]->k = 0;
}
int n, m, k;
int num;
cin >> n >> m;
if (n == 0)
return 0;
Node *n1;
for (int i = 0; i < m; i++)
{
cin >> num >> k;
n1 = nodes[num];
n1->k = k;
for (int j = 0; j < n1->k; ++j)
{
cin >> n1->leaf[j];
}
}
bfs();
for (int i = 0; i < layer; i++)
{
if (i != 0)
cout << " " << leafCount[i];
else
cout << leafCount[i];
}
return 0;
}
感想：
一开始初始化指针数组为NULL，然后输入的时候新建，可能哪个结点输入的时候没建，导致两个样例过不去。dfs的代码写起来好少。
原文地址：https://www.cnblogs.com/stormax/p/10921902.html