HDU 4714 Tree2cycle （树形DP）

Tree2cycle

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 324    Accepted Submission(s): 54

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

 

Recommend

liuyiding

 

简单树形DP来一发

用dp1表示形成一颗链，而且该点在端点。

dp2表示形成一颗链需要的最少步数

 /* *******************************************
 Author       : kuangbin
 Created Time : 2013年09月08日 星期日 12时00分01秒
 File Name    : 1009.cpp
 ******************************************* */
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #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>
 #include <time.h>
 using namespace std;

 const int MAXN = ;
 vector<int>vec[MAXN];
 int dp1[MAXN];
 int dp2[MAXN];
 void dfs(int u,int pre)
 {
     int sz = vec[u].size();
     int sum1 = ;
     int maxn = , maxid = -;
     int smaxn = , smaxid = -;
     for(int i = ;i < sz;i++)
     {
         int v = vec[u][i];
         if(v == pre)continue;
         dfs(v,u);
         sum1 += dp2[v]+;
         int tmp = dp1[v] - (dp2[v] + );
         tmp = -tmp;
         if(tmp > smaxn)
         {
             smaxn = tmp;
             smaxid = v;
             if(smaxn > maxn)
             {
                 swap(smaxn,maxn);
                 swap(smaxid,maxid);
             }
         }
     }
     dp1[u] = sum1 - maxn;
     dp2[u] = sum1 - maxn - smaxn;

 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int n;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         int u,v;
         for(int i = ;i <= n;i++)
             vec[i].clear();
         for(int i = ;i < n;i++)
         {
             scanf("%d%d",&u,&v);
             vec[u].push_back(v);
             vec[v].push_back(u);
         }
         dfs(,-);
         cout<<dp2[]+<<endl;
     }
     return ;
 }