Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9913 | Accepted: 4069 |
Description
For example, consider the tree:
Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two.
For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.
Input
Output
Sample Input
1 7 2 6 1 2 1 4 4 5 3 7 3 1
Sample Output
1 2
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
#define N 21000
#define mem(a,t) memset(a,t,sizeof(a))
const int inf=1000005;
int cnt,n;
struct node
{
int v,next;
}e[N*2];
int head[N];
int num[N];
int ans,index;
void add(int u,int v)
{
e[cnt].v=v;
e[cnt].next=head[u];
head[u]=cnt++;
}
void dfs(int u,int len,int fa)
{
int i,v,tmp=0;
num[u]=1;
for(i=head[u];i!=-1;i=e[i].next)
{
v=e[i].v;
if(v!=fa)
{
dfs(v,len+1,u);
tmp=max(tmp,num[v]);
num[u]+=num[v];
}
}
int t=max(tmp,n-num[u]);
if(t<=ans)
{
if(u<index||t<ans)
{
ans=t;
index=u;
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
int i,u,v,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
cnt=0;
mem(head,-1);
for(i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
mem(num,0);
ans=n;
dfs(1,0,-1);
printf("%d %d\n",index,ans);
}
return 0;
}