PAT (Advanced Level) Practice 1004 Counting Leaves （30 分）
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
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題目大意：給出節點總數和非葉節點總數，求每一層的葉節點個數。
題解1：
思路：
每個節點用一個struct node記錄，node中使用vector<int>child記錄它的子節點。（這題建樹比較簡單）
深度優先搜尋DFS，從根節點01開始，記錄每個節點的層數，遇到葉節點後，該層葉節點總數加1，同時記錄最大層數。
最後輸出所有層數的葉節點個數。
原始碼1：
#include<iostream>
#include<stdlib.h>
#include<vector>
using namespace std;
int n, m, i, j, k, temp,maxlevel=0;
int level[100] = { 0 };
struct node
{
vector<int>child;
};
vector<node>v;
void dfs(int r, int step)
{
int i;
if (v[r].child.size()==0)
{
level[step]++;
if (step > maxlevel)maxlevel = step;
return;
}
for (i = 0; i < v[r].child.size(); i++)
{
dfs(v[r].child[i], step + 1);
}
return;
}
int main()
{
scanf("%d %d", &n, &m);
v.resize(n + 1);
for (i = 1; i <= m; i++)
{
scanf("%d %d", &temp,&k);
v[temp].child.resize(k);
for (j = 0; j < k; j++)
scanf("%d", &v[temp].child[j]);
}
dfs(01, 0);
for (i = 0; i <= maxlevel; i++)
if (i == 0)printf("%d", level[i]);
else printf(" %d", level[i]);
system("pause");
return 0;
}