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。