K - Work 分类: 比赛 2015-07-29 19:13 3人阅读 评论(0) 收藏
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
System Crawler (2015-07-28)
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
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
Sample Input
7 2
1 2
1 3
2 4
2 5
3 6
3 7
Sample Output
2题意就是,有n个人,判断有几个人领导的人数是k,输入两个数,前面的领导后面的;#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <string>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <algorithm> using namespace std; const int MAX = 110; int Map[MAX][MAX]; int Dp[MAX]; int n,k; int DFS(int s)//DFS搜索每个人所领导的人数;
{
if(Dp[s])
{
return Dp[s];
}
int sum=0;
for(int i=1;i<=n;i++)
{
if(Map[s][i])
{
sum++;
sum+=DFS(i);
}
}
Dp[s]=sum;
return Dp[s];
} int main()
{
int a,b;
while(~scanf("%d %d",&n,&k))
{
memset(Map,0,sizeof(Map));
memset(Dp,0,sizeof(Dp));
for(int i=1;i<n;i++)
{
scanf("%d %d",&a,&b);
Map[a][b]=1;
}
int ans=0;
for(int i=1;i<=n;i++)
{
if(DFS(i)==k)
{
ans++;
}
}
printf("%d\n",ans);
}
return 0;
}
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