PAT (Advanced Level) 1004. Counting Leaves (30) 解题报告

PAT (Advanced Level) 1004. Counting Leaves (30) 解题报告

1004. Counting Leaves (30)

时间限制 400 ms
内存限制 65536 kB
代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input

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.

Output

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

分析：题目给出一个家谱，求每一代没有后代的数量

#include <cstdio>
#include <iostream>
#include <map>
#include <vector>
#include <cstring>
#define MAXN 101
#define INF 0x3f3f3f3f
using namespace std;map<int,vector<int> > trees;
int res[MAXN];void dfs(int id, int level)
{if(trees[id].empty()){res[level]++;return;}vector<int>::iterator ite = trees[id].begin();for(;ite != trees[id].end(); ite++)dfs(*ite, level+1);
}int main()
{int n, m, p, k, son, s, cnt;scanf("%d%d", &n, &m);for(int i = 0; i < m; i++){scanf("%d%d", &p, &k);while(k--){scanf("%d", &son);trees[p].push_back(son);}}dfs(1, 0);printf("%d", res[0]);s = n - m;cnt = res[0];for(int i = 1; cnt < s; i++){printf(" %d", res[i]);cnt += res[i];}printf("n");return 0;
}