标签：family line number text ext start math ++ follow
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
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.
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.
2 1
01 1 02
0 1
#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。
标签：family line number text ext start math ++ follow
原文地址：https://www.cnblogs.com/stormax/p/10921902.html
