1004 Counting Leaves (30 分)
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 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
.
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.
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
这个题目的意思就是一个家庭的家族谱,一开始先是辈分最大的人 ,然后下面排着他的孩子,然后就这样一直排到最后的一辈(没有孩子)。这道题的输入是 输入N为总结点的数目,M为非叶子节点的数目,然后以下跟随M行,每行有一个ID K ID[1] ID[2] ... ID[K],id为这个非叶子节点的id,然后K为他的孩子数目,然后k个孩子的id,输入结束。输出:每一个辈分中有几个叶子节点。
光看这个,我们就想到BFS层次遍历寻找叶子节点,找出有多少辈分,然后输出各辈分的叶子节点。
#include<iostream>
#include<queue>
#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
struct node
{
int level;
int num;
int child[maxn];
}dian[maxn];
int shu[maxn];
int solve()
{
memset(shu, 0, sizeof(shu));
queue<node>q;
dian[1].level = 0;
q.push(dian[1]); //一开始根节点就是01 ,就是1
int maxlevel=-1;
while (!q.empty())
{
node t = q.front();
q.pop();
maxlevel = max(maxlevel, t.level);
if (!t.num)
++shu[t.level];
else
{
for (int i = 0; i < t.num; i++)
{
dian[ t.child[i] ].level = t.level + 1;
q.push( dian[t.child[i]] );
}
}
}
return maxlevel;
}
int main()
{
int n, m;
cin >> n >> m;
int id, t;
for (int i = 0; i < m; i++)
{
cin >> id >> t;
dian[id].num = t;
for (int j = 0; j < t; j++)
cin >> dian[id].child[j]; //直接id 当作下标,直接将所有的节点都装进去了
//后面直接插入就行
}
int maxlevel = solve();
for (int i = 0; i <= maxlevel; i++)
{
if (i)
cout << ' ';
cout << shu[i];
}
return 0;
}