hdu 4714 Tree2cycle
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\n",2*ans+1);
}
return 0;
}