1004 Counting Leaves (30point(s))
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
DFS
#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
vector<int> Node[maxn];
int leaf[maxn]={
0};
int maxDepth=0;
void DFS(int index,int depth){
if(Node[index].empty()){
leaf[depth]++;
if(depth>maxDepth){
maxDepth=depth;
}
}
for(int i=0;i<Node[index].size();++i)
DFS(Node[index][i],depth+1);
}
int main(){
int n,m,id,k,child;
scanf("%d%d",&n,&m);
for(int i=1;i<=m;++i){
scanf("%d%d",&id,&k);
for(int j=0;j<k;++j){
scanf("%d",&child);
Node[id].push_back(child);
}
}
DFS(1,1);
for(int i=1;i<=maxDepth;++i){
if(i!=1) printf(" ");
printf("%d",leaf[i]);
}
printf("\n");
}
BFS
#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
struct node{
vector<int> child;
int depth;
}Node[maxn];
int leaf[maxn]={
0};
int maxDepth=0;
void BFS(int index){
queue<int> q;
q.push(index);
Node[index].depth=1;
while(!q.empty()){
int now=q.front();
if(Node[now].child.empty()) leaf[Node[now].depth]++;
if(Node[now].depth>maxDepth) maxDepth=Node[now].depth;
q.pop();
for(int i=0;i<Node[now].child.size();++i){
int child=Node[now].child[i];
Node[child].depth=Node[now].depth+1;
q.push(child);
}
}
}
int main(){
int n,m,id,k,child;
scanf("%d%d",&n,&m);
for(int i=1;i<=m;++i){
scanf("%d%d",&id,&k);
for(int j=0;j<k;++j){
scanf("%d",&child);
Node[id].child.push_back(child);
}
}
BFS(1);
for(int i=1;i<=maxDepth;++i){
if(i!=1) printf(" ");
printf("%d",leaf[i]);
}
printf("\n");
}