PAT甲1004 Counting Leaves【dfs】

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

题意：

给定一棵树和节点之间的关系。要求统计每一层的叶子节点个数。

思路：

就建树，暴力dfs就好了。maxn=105的时候WA和RE了，改成了1005就过了。

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <map>
 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f
 
 const int maxn = ;
 int n, m;
 struct node{
     int v, nxt;
 }edge[maxn];
 int head[maxn], tot = ;
 int cnt[maxn], dep = -;
 
 void addedge(int u, int v)
 {
     edge[tot].v = v;
     edge[tot].nxt = head[u];
     head[u] = tot++;
     edge[tot].v = u;
     edge[tot].nxt = head[v];
     head[v] = tot++;
 }
 
 void dfs(int rt, int fa, int h)
 {
     int sum = ;
     dep = max(dep, h);
     for(int i = head[rt]; i != -; i = edge[i].nxt){
         if(edge[i].v == fa)continue;
         sum++;
         dfs(edge[i].v, rt, h + );
     }
     if(sum == ){
         cnt[h]++;
     }
 
 }
 
 int main()
 {
     scanf("%d%d", &n, &m);
     memset(head, -, sizeof(head));
     for(int i = ; i < m; i++){
         int u, k;
         scanf("%d %d", &u, &k);
         for(int j = ; j < k; j++){
             int v;
             scanf("%d", &v);
             addedge(u, v);
         }
     }
 
     dfs(, -, );
     //cout<<dep<<endl;
     printf("%d", cnt[]);
     for(int i = ; i <= dep; i++){
         printf(" %d", cnt[i]);
     }
     printf("\n");
     return ;
 }