PAT 甲级 1004. Counting Leaves (30) C++版

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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

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

求树中每一层的叶子结点数目，利用广度优先搜索算法，队列保存每一层的节点，遍历队列可以知道哪一些结点没有孩子结点，计数输出即可得到答案。

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

using namespace std;

vector<vector<int>> v(100);

int flag = 0;

void bfs() {
    queue<int> q;
    q.push(1);
    while (!q.empty()) {
        int n = q.size(), cnt = 0;
        for (int i = 0; i < n; i++) {
            int root = q.front();
            q.pop();
            if (v[root].size() == 0) cnt++;
            for (int j = 0; j < v[root].size(); j++) {
                q.push(v[root][j]);
            }
        }
        if (flag) printf(" ");
        flag = 1;
        printf("%d", cnt);
    }
}

int main() {
    int n = 0, m = 0, id = 0, k = 0, t = 0;
    scanf("%d %d", &n, &m);

    for (int i = 0; i < m; i++) {
        scanf("%d %d", &id, &k);
        for (int j = 0; j < k; j++) {
            scanf("%d", &t);
            v[id].push_back(t);
        }
    }
    bfs();
    return 0;
}