2015 Multi-University Training Contest 3 1011(DFS)
Work
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 359 Accepted Submission(s): 255
Problem Description
![](http://acm.hdu.edu.cn/data/images/C581-1011-1.jpg)
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
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
Source
题意:一个公司有N个人,输入管理的顺序,求管理K个人的管理人员有多少人
分析:暴搜
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<string> #include<iostream> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<vector> #include<map> #include<stdlib.h> #include<algorithm> #define LL __int64 using namespace std; const int MAXN=100+5; int n,k,cnt,id; int g[MAXN][MAXN]; int vv[MAXN],cc[MAXN]; int DFS(int cur) { cc[cur]=0; for(int i=1;i<=n;i++) { if(g[cur][i]) cc[cur]+=DFS(i); } if(cc[cur]==k) cnt++; return cc[cur]+1; } int main() { //freopen("in.txt","r",stdin); while(scanf("%d %d",&n,&k)!=EOF) { memset(g,0,sizeof(g)); memset(cc,0,sizeof(cc)); memset(vv,0,sizeof(vv)); for(int i=1;i<n;i++) { int st,en; scanf("%d %d",&st,&en); g[st][en]=1; vv[en]=1; } for(int i=1;i<=n;i++) if(!vv[i]) { id=i; break; } //找出根结点 cnt=0; DFS(id); printf("%d\n",cnt); } return 0; }