1004 Counting Leaves (30 分)
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题目链接:
https://pintia.cn/problem-sets/994805342720868352/problems/994805521431773184
题目描述:
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, 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.
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
题目大意:
一个树,求其每层的叶子结点个数。
思路:
1、通过结构体存储树,将父节点记录。
2、知道了自己的父亲结点的同时也知道了自己的层数(父节点层数+1)。
3、自己是否有孩子结点在输入的时候就可以进行记录(ID即为非叶节点)。
注意事项:
1、输入并非有序,先把所有结点的父节点记录下来,不要直接记录层数,因为可能父节点还没有读取到,就会导致没有层数。
2、当只有一个根时,输出1,根也是一个结点(好吧,是我的树知识不过关~-~)
AC代码:
#include <iostream> #include <cstring> using namespace std; struct { int childHave = 0; //是否为叶子节点 int level; //层数 int father; //父节点 }root[105]; int sum[105]; int main() { memset(sum,0,sizeof(sum)); int n,m; cin >> n >> m; //记录数据 root[1].level = 1; for(int i=1;i<=m;i++) { int r,k; cin >> r >> k; root[r].childHave = 1; for(int j=1;j<=k;j++) { int temp; cin >> temp; root[temp].father = r; } } int maxa = 0; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(root[j].father == i) root[j].level = root[i].level+1; //如果父亲节点为i则层数为i层数+1 if(root[j].level > maxa) maxa = root[j].level; //更新最大层数 } } for(int i=1;i<=n;i++) if(root[i].childHave == 0) sum[root[i].level]++; cout << sum[1]; for(int i=2;i<=maxa;i++) cout << ‘ ‘ << sum[i]; return 0; }
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