Pat Jia 1004 counting leaves [DFS]

1004 counting leaves (30 points)

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]

WhereIDIs a two-digit number representing a given non-leaf node,KIs the number of its children, followed by a sequence of two-digitID'S of its children. For the sake of simplicity, let us fix the root ID to be01.

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, where01Is the root and02Is its only child. Hence on the Root01Level, there is0Leaf node; and on the next level, there is1Leaf node. Then we shoshould output0 1In a line.

Sample input:

2 101 1 02

Sample output:

0 1

 

Question:

Specify the relationship between a tree and a node. Count the number of leaf nodes on each layer.

Ideas:

Just build a new architecture, and it's good to have a violent DFS. When maxn = 105, WA and re are changed to 1005.

 

 1 #include <iostream> 2 #include <set> 3 #include <cmath> 4 #include <stdio.h> 5 #include <cstring> 6 #include <algorithm> 7 #include <vector> 8 #include <queue> 9 #include <map>10 #include <bits/stdc++.h>11 using namespace std;12 typedef long long LL;13 #define inf 0x7f7f7f7f14 15 const int maxn = 1005;16 int n, m;17 struct node{18     int v, nxt;19 }edge[maxn];20 int head[maxn], tot = 0;21 int cnt[maxn], dep = -1;22 23 void addedge(int u, int v)24 {25     edge[tot].v = v;26     edge[tot].nxt = head[u];27     head[u] = tot++;28     edge[tot].v = u;29     edge[tot].nxt = head[v];30     head[v] = tot++;31 }32 33 void dfs(int rt, int fa, int h)34 {35     int sum = 0;36     dep = max(dep, h);37     for(int i = head[rt]; i != -1; i = edge[i].nxt){38         if(edge[i].v == fa)continue;39         sum++;40         dfs(edge[i].v, rt, h + 1);41     }42     if(sum == 0){43         cnt[h]++;44     }45 46 }47 48 int main()49 {50     scanf("%d%d", &n, &m);51     memset(head, -1, sizeof(head));52     for(int i = 0; i < m; i++){53         int u, k;54         scanf("%d %d", &u, &k);55         for(int j = 0; j < k; j++){56             int v;57             scanf("%d", &v);58             addedge(u, v);59         }60     }61 62     dfs(1, -1, 1);63     //cout<<dep<<endl;64     printf("%d", cnt[1]);65     for(int i = 2; i <= dep; i++){66         printf(" %d", cnt[i]);67     }68     printf("\n");69     return 0;70 }

 

Pat Jia 1004 counting leaves [DFS]