标签：from pop node seq roc next while int dep
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
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.
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.
2 1
01 1 02
0 1
数所有叶子节点的个数。先根据数据建树，然后dfs确定每个节点的深度并判断是否为叶节点。我的程序里最后又进行了一次bfs统计各个深度下的叶节点个数，其实可以直接在dfs过程中完成。
#include <iostream> #include <string> #include <map> #include <vector> #include <queue> #include <cstring> using namespace std; struct Node { int depth; vector<Node*> children; Node(int d):depth(d){}; }; map<int,Node*> nodes; int N,M,K; void bfs(Node* ptr,int maxdepth) { queue<Node*> q; q.push(ptr); int ans[maxdepth+1]; memset(ans,0,sizeof ans); int cnt=0; while(!q.empty()) { Node* tmp=q.front(); q.pop(); int sz=tmp->children.size(); if(sz==0) { cnt++; ans[tmp->depth]++; } for(int i=0;i<sz;i++) q.push(tmp->children[i]); } cout<<ans[0]; for(int i=1;i<=maxdepth;i++) cout<<‘ ‘<<ans[i]; cout<<endl; } int maxdepth=0; void dfs(Node* ptr,int depth) { if(ptr==NULL)return; maxdepth=max(depth,maxdepth); ptr->depth=depth; for(int i=0;i<ptr->children.size();i++) { dfs(ptr->children[i],depth+1); } } int main() { cin>>N>>M; if(N==0)return 0; if(M==0){cout<<N<<endl;return 0;} int tid,tid2; for(int i=0;i<M;i++) { cin>>tid>>K; if(nodes.find(tid)==nodes.end()) { nodes[tid]=new Node(0); } for(int j=0;j<K;j++) { cin>>tid2; if(nodes.find(tid2)==nodes.end()) { nodes[tid2]=new Node(0); } nodes[tid]->children.push_back(nodes[tid2]); } } dfs(nodes[1],0); bfs(nodes[1],maxdepth); return 0; }
PAT Advanced 1004 Counting Leaves
标签：from pop node seq roc next while int dep
原文地址：https://www.cnblogs.com/zest3k/p/11443995.html
