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http://acm.hdu.edu.cn/showproblem.php?pid=5326
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.
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
For each test case, output the answer as described above.
7 2
1 2
1 3
2 4
2 5
3 6
3 7
2
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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原文地址：http://www.cnblogs.com/GadyPu/p/4783968.html
