PAT甲1004 Counting Leaves【dfs】

1004 Counting Leaves （30 分）

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

题意：

给定一棵树和节点之间的关系。要求统计每一层的叶子节点个数。

思路：

就建树，暴力dfs就好了。maxn=105的时候WA和RE了，改成了1005就过了。

#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f

const int maxn = 1005;
int n, m;
struct node{
    int v, nxt;
}edge[maxn];
int head[maxn], tot = 0;
int cnt[maxn], dep = -1;

void addedge(int u, int v)
{
    edge[tot].v = v;
    edge[tot].nxt = head[u];
    head[u] = tot++;
    edge[tot].v = u;
    edge[tot].nxt = head[v];
    head[v] = tot++;
}

void dfs(int rt, int fa, int h)
{
    int sum = 0;
    dep = max(dep, h);
    for(int i = head[rt]; i != -1; i = edge[i].nxt){
        if(edge[i].v == fa)continue;
        sum++;
        dfs(edge[i].v, rt, h + 1);
    }
    if(sum == 0){
        cnt[h]++;
    }

}

int main()
{
    scanf("%d%d", &n, &m);
    memset(head, -1, sizeof(head));
    for(int i = 0; i < m; i++){
        int u, k;
        scanf("%d %d", &u, &k);
        for(int j = 0; j < k; j++){
            int v;
            scanf("%d", &v);
            addedge(u, v);
        }
    }

    dfs(1, -1, 1);
    //cout<<dep<<endl;
    printf("%d", cnt[1]);
    for(int i = 2; i <= dep; i++){
        printf(" %d", cnt[i]);
    }
    printf("
");
    return 0;
}