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[]; // 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 = , i;
     while (queSize--) { // traverse all nodes at the current level
         cur = que.front();
         que.pop();

         nodeSize = node[cur].size();
         if (nodeSize == ) { // if it is a leaf node
             count++;
         } else {
             for (i = ; 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() > ) { // if que has elements, traverse
         bfs();
     }
 }

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

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

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

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

     system("pause");
     return ;
 }