hdu 5326 Work

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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. 
   
   
   
   

 

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
2

拓扑排序确定每个结点的权值或者直接树的先序遍历

#include<cstring>
#include<string>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<vector>

using namespace std;

#define INF 0x3f3f3f3f

int s[105];
int in[105];

int e_max;
int fir[5005];
int u[5005],v[5005],nex[5005],w[5005];
inline void init()
{
    e_max=0;
    memset(fir,-1,sizeof fir);
    memset(in,0,sizeof in);
}

inline void add_edge(int s,int t)
{
    int e=e_max++;
    u[e]=s;
    v[e]=t;
    nex[e]=fir[u[e]];
    fir[u[e]]=e;
}

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        init();
        memset(w,0,sizeof w);
        for(int i=0;i<n-1;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            in[a]++;
            add_edge(b,a);
        }
        int left=n;

        int f=0,r=-1;
        int q[1050];
        int vis[1050];
        for(int i=1;i<=n;i++)
        {
            if(in[i]==0)
            {
                in[i]--;
                q[++r]=i;
            }
        }

        while(f<=r)
        {
            int k=q[f++];
            for(int i=fir[k];~i;i=nex[i])
            {
                int e=v[i];
                in[e]--;
                w[e]+=w[k]+1;
                if(in[e]==0)
                {
                    q[++r]=e;
                }
            }
        }

        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(w[i]==k)
                ans++;
        }

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

    }
    return 0;
}
 

#include<cstring>
#include<string>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<vector>

using namespace std;

#define INF 0x3f3f3f3f
#define maxn 105

int n,m;
int w[maxn];
bool vis[maxn];

int fir[maxn];
int u[maxn],v[maxn],nex[maxn];
int e_max;

void init()
{
    e_max=0;
    memset(fir,-1,sizeof fir);
     memset(w,0,sizeof w);
        memset(vis,false,sizeof vis);
}

void add_edge(int s,int t)
{
    int e=e_max++;
    u[e]=s;
    v[e]=t;
    nex[e]=fir[u[e]];
    fir[u[e]]=e;
}

void dfs(int k)
{
    for(int i=fir[k];~i;i=nex[i])
    {
        int e=v[i];
        if(!vis[e])
        {
            vis[e]=true;
            dfs(e);
        }
        w[k]+=w[e]+1;
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
       init();

        for(int i=0;i<n-1;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            add_edge(a,b);
        }

        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i])
            {
                dfs(i);
                vis[i]=true;
            }
            if(w[i]==m) ans++;
        }

        printf("%d\n",ans);
    }
    return 0;
}