PAT-甲级-1004(Counting Leaves )

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博客围绕家族树问题，给定树的节点数、非叶子节点数及父子关系，要求统计从根节点开始每一层的叶子节点数。解题思路是使用广度优先搜索（BFS）从根节点遍历，因是树结构无需标记。

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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

题意： 给定一棵树，树上所有节点的编号都是两位数(01-99)，给定节点之间的父子关系，问你从根节点开始的每一层节点中，有多少个节点无子节点，即：求每一层节点的叶子节点数

解题： 就是bfs从根节点开始跑一遍就完了，因为是树，所以也就不需要标记

#include <bits/stdc++.h>
#define fin freopen("in.txt", "r", stdin)
using namespace std;
typedef long long ll;
const int maxn = 1005;
vector<int> p[maxn];   //存树
int ans[maxn];         //ans[i]表示第i层的叶子节点数
struct node {
    int level, id;
    node(int x, int y) : level(x), id(y) {}
};
int main() {
    int n, m, id, x, k, max_level = 0;
    cin >> n >> m;
    for(int i = 0; i < m; i++) {
        cin >> id >> k;
        for(int j = 0; j < k; j++) { cin >> x; p[id].push_back(x); }
    }
    queue<node> Q;
    Q.push(node(0, 1));
    while(!Q.empty()) {
        node xx = Q.front();
        Q.pop();
        max_level = max(max_level, xx.level);                 //求最大层数
        if((int) p[xx.id].size() == 0) ans[xx.level]++;       //如果该节点是叶子节点，则该层的叶子节点数+1
        for(int i = 0; i < (int) p[xx.id].size(); i++) { 
            Q.push(node(xx.level + 1, p[xx.id][i]));
        }
    }
    for(int i = 0; i <= max_level; i++) if(i) cout << " " << ans[i]; else cout << ans[i];
    puts("");
    return 0;
}