Tree2cycle

Problem Description

A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 unit of cost respectively. The nodes are labeled from 1 to N. Your job is to transform the tree to a cycle(without superfluous edges) using minimal cost.

A cycle of n nodes is defined as follows: (1)a graph with n nodes and n edges (2)the degree of every node is 2 (3) each node can reach every other node with these N edges.

Input

The first line contains the number of test cases T( T<=10 ). Following lines are the scenarios of each test case.
In the first line of each test case, there is a single integer N( 3<=N<=1000000 ) - the number of nodes in the tree. The following N-1 lines describe the N-1 edges of the tree. Each line has a pair of integer U, V ( 1<=U,V<=N ), describing a bidirectional edge (U, V).

Output

For each test case, please output one integer representing minimal cost to transform the tree to a cycle.

Sample Input

1
4
1 2
2 3
2 4

Sample Output

3

Hint

In the sample above, you can disconnect (2,4) and then connect (1, 4) and
(3, 4), and the total cost is 3.

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

#include <stdio.h>
#include <string.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define Maxn 1010000

struct Edge
{
    int v;
    struct Edge *next;
}*head[Maxn<<],edge[Maxn<<];

int n,cnt,ans;
void add(int a,int b)
{
    edge[++cnt].v=b,edge[cnt].next=head[a];
    head[a]=&edge[cnt];
}

int dfs(int cur,int pa) //返回分支个数
{
    int res=;
    struct Edge * p=head[cur];

    while(p)
    {
        if(p->v!=pa) res+=dfs(p->v,cur);
        p=p->next;
    }
    if(res>=) //超过两个分支，将其它分支链过来
    {
        if(cur==) ans+=res-; //树根不用链到其它地方
        else ans+=res-; //选两个，其它的边都要断开和重新链接一次
        return ; //断开了
    }
    else return ; //作为一个单支
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        cnt=;
        memset(head,NULL,sizeof(head));
        for(int i=;i<n;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            add(a,b);
            add(b,a);
        }
        ans=;
        dfs(,);
        printf("%d\n",ans*+);
    }
    return ;
}