题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5326

Work

Description

hdu 5326 Work

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

统计子树大小为k的节点数。。

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<vector>

#include<queue>

#include<set>

using std::set;

using std::pair;

using std::swap;

using std::queue;

using std::multiset;

#define pb(e) push_back(e)

#define sz(c) (int)(c).size()

#define mp(a, b) make_pair(a, b)

#define all(c) (c).begin(), (c).end()

#define iter(c) decltype((c).begin())

#define cls(arr, val) memset(arr, val, sizeof(arr))

#define cpresent(c, e) (find(all(c), (e)) != (c).end())

#define rep(i, n) for(int i = 0; i < (int)n; i++)

#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)

const int N = 110;

const int INF = 0x3f3f3f3f;

struct edge { int to, next; }G[N << 1];

bool inq[N];

int tot, size[N], head[N];

void init(int n) {

    tot = 0;

    rep(i, n + 1) {

        head[i] = -1;

        inq[i] = size[i] = false;

    }

}

inline void add_edge(int u, int v) {

    G[tot].to = v, G[tot].next = head[u], head[u] = tot++;

}

void dfs(int u) {

    size[u] = 1;

    for (int i = head[u]; ~i; i = G[i].next) {

        edge &e = G[i];

        dfs(e.to);

        size[u] += size[e.to];

    }

}

int main() {

#ifdef LOCAL

    freopen("in.txt", "r", stdin);

    freopen("out.txt", "w+", stdout);

#endif

    int n, k, u, v;

    while (~scanf("%d %d", &n, &k)) {

        init(n);

        rep(i, n - 1) {

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

            add_edge(u, v);

            inq[v] = true;

        }

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

            if (!inq[i]) {

                u = i;

                break;

            }

        }

        int sum = 0;

        dfs(u);

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

            if (size[i] == k + 1) sum++;

        }

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

    }

    return 0;

}

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