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.
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
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=0; 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=0; for(int i=0;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>=2) { if(u==1) ans+=2*(sum-2); else ans+=2*(sum-1); 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=0; int from,to; for(int i=0;i<=n;i++) edge[i].clear(); for(int i=0;i<n-1;i++) { scanf("%d%d",&from,&to); //cout<<from<<" "<<to<<endl; edge[from].push_back(to); edge[to].push_back(from); }//建树 dfs(1,-1); printf("%d ",ans+1); } return 0; }