标签：
Tree2cycle
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1730 Accepted Submission(s): 401
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
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
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具体思路http://blog.csdn.net/dongdongzhang_/article/details/11395305
额，，不能算dp吧，充其量算个树上的乱搞。。会爆栈。。开栈就过了
ac代码
#include<stdio.h>
#include<string.h>
#pragma comment(linker, "/STACK:102400000,102400000")
int head[1000010],vis[1000010],cnt,ans;
struct s
{
int u,v,next;
}edge[1000010<<1];
void add(int u,int v)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
int dfs(int u)
{
vis[u]=1;
int sum=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(!vis[v])
{
sum+=dfs(v);
}
}
if(sum>=2)
{
if(u==1)
ans+=(sum-2)*2;
else
ans+=(sum-1)*2;
return 0;
}
return 1;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
cnt=0;
ans=0;
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
scanf("%d",&n);
int i;
for(i=1;i<n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
add(a,b);
add(b,a);
}
dfs(1);
printf("%d\n",ans+1);
}
}
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HDOJ 题目4714 Tree2cycle（树形DP）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/47405779
