1004_Counting_Leaves

1004. Counting Leaves (30)

时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input

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.
Output

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

-----------------------------------------------------------------------------------------------------------------
解：每个非叶子结点有若干个子结点，但对于树来说，每个结点只有一个父结点，这样存储感觉清晰一些

#include
using namespace std;
 
int parent[101];
int leaf[101];
int level[101];
int maxlevel = 1;
 
//用一个parent[]保存父结点，而不是老想用一个数组保存子结点
void findLevel(int x){
    int r = x;
    int temp = 1;
    while(parent[r] != r)
    {
        temp++;
        r = parent[r];
    }
    level[temp]++;
    if(temp > maxlevel)
        maxlevel = temp;
}
int main()
{
    int node_num, non_leaf_node_num;
    int parent_id, children_num, child_id;
    cin >> node_num >> non_leaf_node_num;
 
    for (int i = 0; i < 101; ++i)
    {
        leaf[i] =  1;
        parent[i] = i;   //每个结点id唯一，因此这样赋值能区别开
        level[i] = 0;
    }
 
    for (int i = 0; i < non_leaf_node_num; ++i)  {       cin >> parent_id >> children_num;
        for(int j = 0; j < children_num; ++j)        {           cin >> child_id;
            parent[child_id] = parent_id;
        }
        leaf[parent_id] = 0;
    }
 
    for(int i = 1; i        if(leaf[i] == 1)
            findLevel(i);
    }
    cout<<level[1];
    for(int i = 2; i        {
            cout << ' '<<level[i];
        }
    return 0;
}