Tree2cycle
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 748 Accepted Submission(s): 172
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
这题 我们主要,根据一点,如果,一个树,它的子树,有一个是超过2个结点的,我们就可以把这个子树,分离开来,子树化成一条直线,最终把所有的小直线相连,就可以得到一个环,而且,得到的一定 是最小值 !
#pragma comment(linker,"/STACK:1024000000,1024000000") #include <iostream> #include <stdio.h> #include <vector> #include <string.h> using namespace std; #define MAXN 1000050 int visit[MAXN],ans; vector<int> vec[MAXN]; int dfs(int u) { visit[u]=1; int i,res=0; for(i=0;i<vec[u].size();i++) { if(!visit[vec[u][i]]) res+=dfs(vec[u][i]); } if(res>=2) { if(u==1) ans+=res-2; else ans+=res-1; return 0; } else return 1; } int main() { int n,i,tcase,s,e; scanf("%d",&tcase); while(tcase--) { scanf("%d",&n); for(i=1;i<=n;i++) vec[i].clear(); for(i=1;i<n;i++) { scanf("%d%d",&s,&e); vec[s].push_back(e); vec[e].push_back(s); } ans=0; memset(visit,0,sizeof(visit)); dfs(1); printf("%d ",2*ans+1); } return 0; }