PAT 1004 Counting Leaves

1004 Counting Leaves

分数 30

作者 CHEN, Yue

单位 浙江大学

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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

栈限制

8192 KB

dfs写法：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n,m,id,k,son;
vector<int> v[105];
int leaf[105];
int maxdepth;void dfs(int id,int depth){if(v[id].size()==0){leaf[depth]++;maxdepth=max(maxdepth,depth);return ;}for(int i=0;i<v[id].size();i++){dfs(v[id][i],depth+1);}
}
int main(){scanf("%d%d",&n,&m);while(m--){scanf("%d%d",&id,&k);while(k--){scanf("%d",&son);v[id].push_back(son);}}dfs(1,1);// id为1的结点（根节点）深度为1printf("%d",leaf[1]);for(int i=2;i<=maxdepth;i++){printf(" %d",leaf[i]);}printf("\n");return 0;
}

bfs写法：

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int N=105;
int n,m;
vector<int> v[N];
int leaf[N];
int maxdepth;
typedef struct{int id;int depth;
}node;
queue<node> q;void bfs(){while(q.size()){node n=q.front();// n为队首结点int id=n.id;int depth=n.depth;if(v[id].size()==0){leaf[depth]++;maxdepth=max(maxdepth,depth);}for(int i=0;i<v[id].size();i++){q.push({v[id][i],depth+1});}q.pop();}
}
int main(){scanf("%d%d",&n,&m);while(m--){int id,k;scanf("%d%d",&id,&k);while(k--){int son;scanf("%d",&son);v[id].push_back(son);}}q.push({1,1});// 根节点的层数为1bfs();printf("%d",leaf[1]);for(int i=2;i<=maxdepth;i++){printf(" %d",leaf[i]);}printf("\n");return 0;
}