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.
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
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
1 /* ******************************************* 2 Author : kuangbin 3 Created Time : 2013年09月08日 星期日 12时00分01秒 4 File Name : 1009.cpp 5 ******************************************* */ 6 #pragma comment(linker, "/STACK:1024000000,1024000000") 7 #include <stdio.h> 8 #include <algorithm> 9 #include <iostream> 10 #include <string.h> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 21 const int MAXN = 1000010; 22 vector<int>vec[MAXN]; 23 int dp1[MAXN]; 24 int dp2[MAXN]; 25 void dfs(int u,int pre) 26 { 27 int sz = vec[u].size(); 28 int sum1 = 0; 29 int maxn = 0, maxid = -1; 30 int smaxn = 0, smaxid = -1; 31 for(int i = 0;i < sz;i++) 32 { 33 int v = vec[u][i]; 34 if(v == pre)continue; 35 dfs(v,u); 36 sum1 += dp2[v]+2; 37 int tmp = dp1[v] - (dp2[v] + 2); 38 tmp = -tmp; 39 if(tmp > smaxn) 40 { 41 smaxn = tmp; 42 smaxid = v; 43 if(smaxn > maxn) 44 { 45 swap(smaxn,maxn); 46 swap(smaxid,maxid); 47 } 48 } 49 } 50 dp1[u] = sum1 - maxn; 51 dp2[u] = sum1 - maxn - smaxn; 52 53 } 54 int main() 55 { 56 //freopen("in.txt","r",stdin); 57 //freopen("out.txt","w",stdout); 58 int T; 59 int n; 60 scanf("%d",&T); 61 while(T--) 62 { 63 scanf("%d",&n); 64 int u,v; 65 for(int i = 1;i <= n;i++) 66 vec[i].clear(); 67 for(int i = 1;i < n;i++) 68 { 69 scanf("%d%d",&u,&v); 70 vec[u].push_back(v); 71 vec[v].push_back(u); 72 } 73 dfs(1,-1); 74 cout<<dp2[1]+1<<endl; 75 } 76 return 0; 77 }