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題目
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
雖然是30分的題，但是好像比1003還簡單。
同樣使用了鄰接鏈表（adjacent list），數據結構如下：
- Member（Vertex）：表示一個家庭成員，我使用了兩個結構成員：
- level，表示這個人在家譜中的輩分，最高的輩分是0，輩分越低，level越大
- child，指向Child結構變量，即鏈表的第一個節點
- Child（Edge）：表示親子關系，也使用了兩個結構成員：
- ID，表示這個孩子的ID
- iter，指向相同父母的下一個孩子Child變量
讀取數據，建立鄰接鏈表。初始化01節點的level為0，其他level為INF。
然后按level從0開始，依次遍歷所有Member，找出相應輩分的家庭成員。這時分兩種情況：這個成員沒有孩子，那么需要進行計數；這個成員有孩子，那么需要更新他們的level值為這個成員的level+1，以便下一次遍歷時找到他們。
使用一個標記表明何時不需再遍歷，我用的是記錄人數，當之前所有level的人數已經是總人數，那么說明已全部遍歷，此時退出循環。
代碼
最新代碼@github，歡迎交流
#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;
}
