A family hierarchy is usually PResented by a pedigree tree. Your job is to count those family members who have no child.InputEach 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.OutputFor 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 Input2 101 1 02Sample Output0 1#include <iostream>#include <cstdio>#include <cstring>#define N 105struct node { int child[N]; int k;}s[N];int que[N*N];void init() { int i; for (i = 0; i < N; i++) s[i].k = 0;}int main() { int n,m,i,j,l,r,count,k,t; while (scanf("%d%d",&n,&m) != EOF) { init(); for (t = 0; t < m; t++) { scanf("%d",&i); scanf("%d",&s[i].k); for (j = 0; j < s[i].k; j++) { scanf("%d",&s[i].child[j]); } } que[0] = 1;//根節點的編號是1 k = 1;//k表示當前的節點的子節點（孩子節點） //l for (l = 0, r = 1,t = 0; l < k; r = k,t++) { count = 0; for (; l < r; l++) { i = que[l]; if (!s[i].k) count++; for (j = 0; j < s[i].k; j++) { que[k++] = s[i].child[j]; } } if (t) printf(" %d",count); else printf("%d",count); } printf("/n"); } return 0;}
