1004 Counting Leaves (30 分)
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<iostream>
#include<algorithm>
#include<vector>
#include<math.h>
using namespace std;
struct Node{
int deep;
vector<int> child;
}node[110];
int n, m, d[110] = { 0 }, h = 0;
void dfs(int root, int deep){
node[root].deep = deep;
h = max(h, deep);
if (node[root].child.size() == 0){
d[deep]++;
return;
}
for (int i = 0; i < node[root].child.size(); i++)
dfs(node[root].child[i], deep + 1);
}
int main(){
scanf("%d %d", &n, &m);
int no, k, x;
for (int i = 0; i < m; i++){
scanf("%d %d", &no, &k);
for (int j = 0; j < k; j++){
scanf("%d", &x);
node[no].child.push_back(x);
}
}
dfs(1, 0);
for (int i = 0; i <= h;i++)
if (i != h)
printf("%d ", d[i]);
else
printf("%d", d[i]);
return 0;
}