hdu 4714 Tree2cycle

Tree2cycle

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 792    Accepted Submission(s): 182

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

141 22 32 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.

 

 

#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>using namespace std;const int MAXN = 1000000 + 50;struct Edge{int v, next;}edge[MAXN * 2];bool vis[MAXN];int head[MAXN], e, ans, n;void add(int u, int v){edge[e].v = v;edge[e].next = head[u];head[u] = e++;}void init(){e = 0;ans = 0;memset(head, -1, sizeof(head));memset(vis, false, sizeof(vis));}bool dfs(int x){int sum = 0;vis[x] = true;for (int i = head[x]; i != -1; i = edge[i].next){int v = edge[i].v;if (!vis[v]){sum += dfs(v);}}if (sum >= 2){ans += x == 1 ? 2 * (sum - 2) : 2 * (sum - 1);return false;}else{return true;}}void input(){int t, u, v;cin >> t;while (t--){cin >> n;init();for (int i = 0; i < n - 1; i++){cin >> u >> v;add(u, v);add(v, u);}dfs(1);cout << ans + 1 << endl;}}int main(){std::ios::sync_with_stdio(false);input();return 0;}