PAT 1004 Counting Leaves

PAT 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 containing0<N<100, the number of nodes in a tree, andM(<N), the number of non-leaf nodes. ThenMlines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

whereIDis a two-digit number representing a given non-leaf node,Kis the number of its children, followed by a sequence of two-digitID's of its children. For the sake of simplicity, let us fix the root ID to be01.

The input ends withNbeing 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no childfor every seniority levelstarting 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, where01is the root and02is its only child. Hence on the root01level, there is0leaf node; and on the next level, there is1leaf node. Then we should output0 1in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

注意点

本题的意思就是输出每一层叶子结点的个数。
注意最大深度max_h的记录，在dfs内部直接去最大值即可，便于后面输出。
输出时由于格式问题，先输出第一个（特殊），在输出之后几个（一般）更加简便。
dfs相对于bfs比较简单一点，无需新开一个数组记录结点的高度。

代码

深度优先遍历（dfs）

#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 110;

struct node
{
    vector<int> child;
}Node[maxn];
int n, m;
int num[maxn];

int max_h = 0;
void dfs(int index, int depth)
{
    max_h = max(depth, max_h);
    if (Node[index].child.size() == 0)
    {
        num[depth] ++;
        return;
    }
    
    for (int i = 0; i < Node[index].child.size(); i ++)
    {
        dfs(Node[index].child[i], depth + 1);
    }
}

int main()
{
    scanf("%d %d", &n, &m);
    int child;
    int id, k;
    for (int i = 1; i <= m; i ++)
    {
        scanf("%d %d", &id, &k);
        for (int j = 0; j < k; j ++)
        {
            scanf("%d", &child);
            Node[id].child.push_back(child);
        }
    }
    
    dfs(1, 1);
    printf("%d", num[1]);
    
    for (int i = 2; i <= max_h; i ++)
    {
        printf(" %d", num[i]);
    }
    
    return 0; 
}

广度优先遍历（bfs）

#include <cstdio>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn = 110;

struct node
{
    vector<int> child;
}Node[maxn];

int n, m;
int num[maxn];
int h[maxn];    //记录结点的高度 
int max_h = 0;

void bfs(int root)
{
    queue<int> q;
    q.push(root);
    while (!q.empty())
    {
        int now = q.front();
        q.pop();
        max_h = max(max_h, h[now]);
        if (Node[now].child.size() == 0)
        {
            num[h[now]] ++;
        }
        for (int i = 0; i < Node[now].child.size(); i ++)
        {
            int child = Node[now].child[i];
            h[child] = h[now] + 1;
            q.push(child);
        }
    }
}

int main()
{
    scanf("%d %d", &n, &m);
    int child;
    int id, k;
    for (int i = 1; i <= m; i ++)
    {
        scanf("%d %d", &id, &k);
        for (int j = 0; j < k; j ++)
        {
            scanf("%d", &child);
            Node[id].child.push_back(child);
        }
    }
    
    h[1] = 1;
    bfs(1);
    printf("%d", num[1]);
    for (int i = 2; i <= max_h; i ++)
    {
        printf(" %d", num[i]);
    }
    
    return 0; 
}

PAT 1004 Counting Leaves