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

[align=left]Problem Description[/align]

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.

[align=left]Input[/align]

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

[align=left]Output[/align]

For each test case, output the answer as described above.

[align=left]Sample Input[/align]

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;
}