A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input Specification:
The input consists of several test cases, each starts with a line containing 0, the number of nodes in a tree, and M (<), 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.
For example, the first 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
1 0
7 4
01 2 02 03
06 1 07
02 2 04 05
03 1 06
0 0
Sample Output:
0 1 1 0 0 2 1
代码:
#include <stdio.h> #include <stdlib.h> struct Node { int index; int c[101]; }s[101]; int leave[101]; int mheight; void count(int id,int height) { if(mheight < height)mheight = height; if(s[id].index == 0) { leave[height] ++; return; } for(int i = 0;i < s[id].index;i ++) count(s[id].c[i],height + 1); } int main() { int n,m,k,id,d; while(scanf("%d%d",&n,&m) && n) { mheight = 0; for(int i = 0;i <= n;i ++) { leave[i] = s[i].index = 0; } for(int i = 0;i < m;i ++) { scanf("%d%d",&id,&k); s[id].index = 0; while(k --) { scanf("%d",&d); s[id].c[s[id].index ++] = d; } } count(1,0); printf("%d",leave[0]); for(int i = 1;i <= mheight;i ++) printf(" %d",leave[i]); putchar(' '); } }