Tree2cycle
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 2161 Accepted Submission(s): 508
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).
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
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
Recommend
/*
每个节点如果有两个或两个以上结点，就将它先于父节点断开，然后在与子节点断开，全部断开之后，和子节点自称一条直链，最后将所有的子链连城一条
直线一个结点有sum个子结点，断开是需要sum-2次操作，连接时需要sum-2次操作，这样一个结点就会进行2*（sum-2）次操作，这时加上断开父节点，一共
进行了2*（sum-2）+1次操作，构成直链之后还需要和父节点连接，这样就一共是2*（sum-1）次操作，如果这个结点是刚开始的结点，那么就不用考虑父节
点，这样操操作就是2*（sum-2）次
*/
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#define N 1000009
using namespace std;
int t,n;
vector<int > edge[N];
int ans=;
bool dfs(int u,int p)//返回你搜到的是不是u的子树
{
//cout<<"ans="<<ans<<endl;
int sum=;
for(int i=;i<edge[u].size();i++)
{
int v=edge[u][i];//下一步
if(v==p) continue;//下一步走过了，就不要走了
sum+=dfs(v,u);
//visit[v]=true;
//cout<<sum<<endl;
}
if(sum>=)
{
if(u==)
ans+=*(sum-);
else
ans+=*(sum-);
return false;
}
return true;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
ans=;
int from,to;
for(int i=;i<=n;i++)
edge[i].clear();
for(int i=;i<n-;i++)
{
scanf("%d%d",&from,&to);
//cout<<from<<" "<<to<<endl;
edge[from].push_back(to);
edge[to].push_back(from);
}//建树
dfs(,-);
printf("%d\n",ans+);
}
return ;
}