HDU 5326 Work (基礎樹形dp)
Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27 Accepted Submission(s): 19
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.
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
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
2015 Multi-University Training Contest 3
題目連結:http://acm.hdu.edu.cn/showproblem.php?pid=5326
題目大意:給一個樹,點之間是僱傭關係 A B表示A是B上司,問有幾個人有k個下屬
題目分析:多校簽到,隨便算一下
題目連結:http://acm.hdu.edu.cn/showproblem.php?pid=5326
題目大意:給一個樹,點之間是僱傭關係 A B表示A是B上司,問有幾個人有k個下屬
題目分析:多校簽到,隨便算一下
#include <cstdio>
#include <cstring>
int const MAX = 105;
struct EDGE
{
int v, next;
}e[MAX];
int head[MAX], num[MAX];
bool in[MAX], vis[MAX];
int cnt, ans;
int n, k;
void Add(int u, int v)
{
e[cnt].v = v;
e[cnt].next = head[u];
head[u] = cnt ++;
}
void DFS(int u)
{
vis[u] = true;
num[u] = 0;
for(int i = head[u]; i != -1; i = e[i].next)
{
int v = e[i].v;
if(!vis[v])
{
DFS(v);
num[u] += (num[v] + 1);
}
}
if(num[u] == k)
ans ++;
return;
}
int main()
{
while(scanf("%d %d", &n, &k) != EOF)
{
cnt = 0;
ans = 0;
memset(head, -1, sizeof(head));
memset(in, false, sizeof(in));
memset(vis, false, sizeof(vis));
for(int i = 0; i < n - 1; i++)
{
int u, v;
scanf("%d %d", &u, &v);
Add(u, v);
in[v] ++;
}
for(int i = 1; i <= n; i++)
if(in[i] == 0)
DFS(i);
printf("%d\n", ans);
}
}
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