Tree2cycle HDU - 4714 (树形dp)

最新推荐文章于 2020-04-26 18:44:28 发布

untilyouydc

最新推荐文章于 2020-04-26 18:44:28 发布

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分类专栏：树形DP

本文链接：https://blog.youkuaiyun.com/qq_40774175/article/details/81436927

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本文介绍了一种算法，用于解决给定一棵N个节点的树，如何以最小的成本将其转化为一个环的问题。通过深度优先搜索计算分支数目，并最终得出转换所需的最小成本。

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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

题意：题目的意思转化一下就是让你求分支树*2+1

如何求分支数？直接考虑每个子树对答案的贡献就行（注意根节点要选择2条作为非分支，因为这样可以保证主干最长）

代码：（友情提示，交c++）

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<vector>
#include<algorithm>
#include<set>
#include<sstream>
#include<cstdio>
#include<unordered_map>
#include<unordered_set>
#include<cmath>
#include<climits>
using namespace std;
const int maxn=2e6+9;
int head[maxn];
int num;
int ans;
int n,m;
struct Edge
{
    int u,v,w,next;
}edge[maxn];
void addEdge(int u,int v,int w)
{
    edge[num].u=u;
    edge[num].v=v;
    edge[num].w=w;
    edge[num].next=head[u];
    head[u]=num++;
}
void init()
{
    memset(head,-1,sizeof(head));
    ans=0;
    num=0;
}
int dfs(int u,int fa)
{ 
    int tmp=0;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==fa)  continue;
        tmp+=dfs(v,u);

    }
    if(tmp>=2)
    {
        if(u==1)
        {
            ans+=tmp-2;
        }
        else
        {
            ans+=tmp-1;
        }
        return 0;
    }
    else
    {
        return 1;
    }
}
int main(int argc, char const *argv[])
{
    int T;
    cin>>T;
    while(T--)
    {
        init();
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addEdge(u,v,0);
            addEdge(v,u,0);
        }
        dfs(1,-1);
        printf("%d\n",ans*2+1);
    }
    return 0;
}