題干：

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

題目大意：

給你一棵樹，有N個節點和M個非葉子結點（N,M<100），已知所有非葉子節點的編號和對應的孩子節點，讓你輸出每一個深度有多少個葉子節點。

解題報告：

按照輸入建樹，然后直接dfs預處理深度，On遍歷得出答案即可。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5;vector<int> tr[222]; int dep[222],ok[222],ans[MAX],n,m; void dfs(int cur,int fa) {dep[cur] = dep[fa] + 1;int up = tr[cur].size();for(int i = 0; i<up; i++) {int v = tr[cur][i];dfs(v,cur);} } int main() {cin>>n>>m;for(int tmp,id,k,i = 1; i<=m; i++) {scanf("%d%d",&id,&k);ok[id] = 1;for(int j = 1; j<=k; j++) {scanf("%d",&tmp); tr[id].push_back(tmp);}}dfs(1,0);int mx = 0;for(int i = 1; i<=n; i++) {mx = max(mx,dep[i]);if(ok[i] == 1) continue;ans[dep[i]]++;}for(int i = 1; i<=mx; i++) {printf("%d%c",ans[i],i == mx ? '\n' : ' '); }return 0; }

?

總結

以上是生活随笔為你收集整理的【PAT - 甲级1004】Counting Leaves (30分) （dfs，递归）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。