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题目
A family hierarchy is usually presented by a pedigree tree. Your job is to
count those family members who have no child.
Input Specification:
Each input file contains one test case. Each case starts with a line
containing , the number of nodes in a tree, and
(
), the
number of non-leaf nodes. Then 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 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.
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
思路
虽然是30分的题，但是好像比1003还简单。
同样使用了邻接链表（adjacent list），数据结构如下：
- Member（Vertex）：表示一个家庭成员，我使用了两个结构成员：
- level，表示这个人在家谱中的辈分，最高的辈分是0，辈分越低，level越大
- child，指向Child结构变量，即链表的第一个节点
- Child（Edge）：表示亲子关系，也使用了两个结构成员：
- ID，表示这个孩子的ID
- iter，指向相同父母的下一个孩子Child变量
读取数据，建立邻接链表。初始化01节点的level为0，其他level为INF。
然后按level从0开始，依次遍历所有Member，找出相应辈分的家庭成员。这时分两种情况：这个成员没有孩子，那么需要进行计数；这个成员有孩子，那么需要更新他们的level值为这个成员的level+1，以便下一次遍历时找到他们。
使用一个标记表明何时不需再遍历，我用的是记录人数，当之前所有level的人数已经是总人数，那么说明已全部遍历，此时退出循环。
代码
最新代码@github，欢迎交流
#include <stdio.h>
#define MAX 999
typedef struct Member *Member;
typedef struct Child *Child;
struct Member{
int level;
Child child;
};
struct Child{
int ID;
Child iter;
};
int main()
{
int N, M, ID, cID, K;
struct Member nodes[100];
struct Child children[100];
/* Read data and initiate a adjacent list */
scanf("%d %d", &N, &M);
for(int i = 1; i <= N; i++)
{
nodes[i].level = MAX;
nodes[i].child = NULL;
}
nodes[1].level = 0; /* root node at level 0 */
for(int i = 0, k = 0; i < M; i++)
{
scanf("%d %d", &ID, &K);
for(; K--; k++)
{
scanf("%d", &cID);
children[k].ID = cID;
children[k].iter = nodes[ID].child;
nodes[ID].child = &children[k];
}
}
/* For every level, find leaf nodes */
int n = N, count;
for(int level = 0; n; level++)
{
count = 0;
for(int i = 1; i <= N; i++) if(nodes[i].level == level)
{
n--;
if(nodes[i].child == NULL)
count++;
/* set the children to next level */
for(Child c = nodes[i].child; c; c = c->iter)
nodes[c->ID].level = level + 1;
}
printf("%d%c", count, n ? ' ' : '\0');
}
return 0;
}
