PAT-A1004. Counting Leaves (30)

　　根据家谱树从根结点开始输出每一层的叶子结点数量。使用BFS来解决。因为不会重复访问结点，所以不需要vis数组来标记是否访问过该结点。

//#include "stdafx.h"
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

vector<int> node[101]; // Storing children's dynamic arrays
queue<int> que; // que for bfs
bool first = true; // first output flag

void bfs() { // traverse by layer
    int queSize = que.size(), cur, nodeSize, count = 0, i;
    while (queSize--) { // traverse all nodes at the current level
        cur = que.front();
        que.pop();

        nodeSize = node[cur].size();
        if (nodeSize == 0) { // if it is a leaf node
            count++;
        } else {
            for (i = 0; i < nodeSize; i++) { // if not, add child node
                que.push(node[cur][i]);
            }
        }
    }

    // output
    if (first) {
        first = false;
    } else {
        printf(" ");
    }
    printf("%d", count);

    if (que.size() > 0) { // if que has elements, traverse
        bfs();
    }
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);

    int i, id, k, child, j;
    for (i = 0; i < m; i++) {
        scanf("%d%d", &id, &k);

        for (j = 1; j <= k; j++) {
            scanf("%d", &child);
            node[id].push_back(child);
        }
    }

    que.push(1); // add root node
    bfs();
    printf("
");

    system("pause");
    return 0;
}