hdu4714树形DP+贪心(乱搞)

Tree2cycle

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.

InputThe 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). 
OutputFor 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.
 
题解：
　　

　　

题解：
　　对于子节点，如果有两个子节点，那么就要分开，分开子孙和祖先是一样的。
　对于根节点，取两个子孙连起来，将其它节点分开。

 #include<iostream>
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #define N 1000007
 using namespace std;
 
 int tot,head[N],Next[N*],info[N*];
 int dp[N];
 
 void init()
 {
     memset(head,-,sizeof head);
     memset(dp,,sizeof dp);
     tot=;
 }
 void add(int fr,int to)
 {
     Next[tot]=head[fr];
     info[tot]=to;
     head[fr]=tot++;
 }
 bool dfs(int u,int f)
 {
     int ans=;
     for(int i=head[u];i!=-;i=Next[i])
     {
         int v=info[i];
         if(v==f) continue;
         if(dfs(v,u)) dp[u]+=dp[v]+;
         else
         {
                dp[u]+=dp[v];
             ans++;
         }
     }
     if(ans==||ans==) return false;
        if(ans>) dp[u]+=ans-;
     return true;
 }
 int main()
 {
     int cas;
     scanf("%d",&cas);
     while(cas--)
     {
         int n;
         scanf("%d",&n);
         init();
         for(int i=,x,y=;i<n;i++)
         {
             scanf("%d%d",&x,&y);
             add(x,y),add(y,x);
         }
         dfs(,-);
         printf("%d\n",dp[]*+);
     }
 }