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*2],info[N*2];
int dp[N];

void init()
{
    memset(head,-1,sizeof head);
    memset(dp,0,sizeof dp);
    tot=0;
}
void add(int fr,int to)
{
    Next[tot]=head[fr];
    info[tot]=to;
    head[fr]=tot++;
}
bool dfs(int u,int f)
{
    int ans=0;
    for(int i=head[u];i!=-1;i=Next[i])
    {
        int v=info[i];
        if(v==f) continue;
        if(dfs(v,u)) dp[u]+=dp[v]+1;
        else
        {
               dp[u]+=dp[v];
            ans++;
        }
    }
    if(ans==0||ans==1) return false;
       if(ans>2) dp[u]+=ans-2;
    return true;
}
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        int n;
        scanf("%d",&n);
        init();
        for(int i=1,x,y=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y),add(y,x);
        }
        dfs(1,-1);
        printf("%d
",dp[1]*2+1);
    }
}