ACM学习历程—HDU 5326 Work（树形递推）

Problem Description

It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company.
As is known to all, every stuff in a company has a title, everyone except the boss has a direct leader, and all the relationship forms a tree. If A’s title is higher than B(A is the direct or indirect leader of B), we call it A manages B.
Now, give you the relation of a company, can you calculate how many people manage k people.

 

 

Input

There are multiple test cases.
Each test case begins with two integers n and k, n indicates the number of stuff of the company.
Each of the following n-1 lines has two integers A and B, means A is the direct leader of B.

1 <= n <= 100 , 0 <= k < n
1 <= A, B <= n

 

 

Output

For each test case, output the answer as described above.

 

 

Sample Input

7 2

1 2

1 3

2 4

2 5

3 6

3 7

 

Sample Output

2

这个题目是个递推，不过由于是树形的，需要dfs来完成递推的过程。

关键在于p[now] += p[to]+1;如果now能manage to的话。

此处采用链式前向星来保存关系图。

 

代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <algorithm>

#define LL long long



using namespace std;



const int maxN = 105;



struct Edge

{

    int to, next;

}edge[maxN];



int head[maxN], cnt;



void addEdge(int u, int v)

{

    edge[cnt].to = v;

    edge[cnt].next = head[u];

    head[u] = cnt;

    cnt++;

}



void initEdge()

{

    memset(head, -1, sizeof(head));

    cnt = 0;

}



int n, k;

int fa[maxN], p[maxN];



void input()

{

    initEdge();

    memset(p, -1, sizeof(p));

    int u, v;

    for (int i = 1; i < n; ++i)

    {

        scanf("%d%d", &u, &v);

        addEdge(u, v);

    }

}





void dfs(int now)

{

    p[now] = 0;

    int to;

    for (int i = head[now]; i != -1; i = edge[i].next)

    {

        to = edge[i].to;

        if (p[to] == -1)

            dfs(to);

        p[now] += p[to]+1;

    }

}



void work()

{

    int ans = 0;

    for (int i = 1; i <= n; ++i)

    {

        if (p[i] != -1)

        {

            if (p[i] == k)

                ans++;

        }

        else

        {

            dfs(i);

            if (p[i] == k)

                ans++;

        }

    }

    printf("%d\n", ans);

}



int main()

{

    //freopen("test.in", "r", stdin);

    while (scanf("%d%d", &n, &k) != EOF)

    {

        input();

        work();

    }

    return 0;

}