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
题意:
给一棵树,寻找每一层的叶子节点数。输入的第一行的两个数是结点的总数和非叶子结点的个数。
Code:
#include<iostream> #include<vector> #include<algorithm> #include<cstdio> using namespace std; vector<int> v[100]; int book[100], maxdepth = -1; void dfs(int index, int depth) { if (v[index].size() == 0) { //是不是叶子节点 book[depth]++; maxdepth = max(maxdepth, depth); return ; } for (int i = 0; i < v[index].size(); ++i) // 遍历该节点的子节点 dfs(v[index][i], depth + 1); } int main() { int n, m, k, node, c; scanf("%d %d", &n, &m); for (int i = 0; i < m; ++i) { scanf("%d %d", &node, &k); for (int j = 0; j < k; ++j) { scanf("%d", &c); v[node].push_back(c); } } dfs(1, 0); printf("%d", book[0]); for(int i = 1; i <= maxdepth; ++i) printf(" %d", book[i]); return 0; }
参考: