hdu 4714 Tree2cycle(简单图或树DP，4级)

Tree2cycle

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 354    Accepted Submission(s): 65

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

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

2013 ACM/ICPC Asia Regional Online —— Warmup

 

Recommend

liuyiding

 思路：居然还真是像我想的那样，凡是碰到多于两个分支的全砍掉，砍掉的分支*2+1就是答案。

#include<cstdio>
#include<cstring>
#include<iostream>
#pragma comment(linker,"/STACK:102400000,102400000")
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=2e6+9;
int head[mm],edge,n;
class Edge{
 public:int v,next;
}e[mm];
void data()
{
  clr(head,-1);edge=0;
}
void add(int u,int v)
{
  e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
}
int ans;
int dfs(int u,int fa)
{ int v;
  int id=0;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(v==fa)continue;
    id+=dfs(v,u);
  }
  if(id>=2)
  {
    if(u==1)//root;
      ans+=id-2;
    else ans+=id-1;
    return 0;
  }
  else return 1;//不切
}
void find_bcc()
{ ans=0;
  dfs(1,-1);
  printf("%d\n",ans+ans+1);
}
int main()
{
  int cas;
  while(~scanf("%d",&cas))
  { int a,b;
    while(cas--)
    { data();
      scanf("%d",&n);
      FOR(i,2,n)
      {
        scanf("%d%d",&a,&b);add(a,b);add(b,a);
      }
      find_bcc();
    }
    
  }
}

dp

#include<cstdio>
#include<cstring>
#include<iostream>
#pragma comment(linker,"/STACK:102400000,102400000")
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=2e6+9;
int head[mm],edge,n;
int dp[mm][2];
class Edge{
 public:int v,next;
}e[mm];
void data()
{
  clr(head,-1);edge=0;
}
void add(int u,int v)
{
  e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
}
int ans;
void dfs(int u,int fa)
{ int v;
  dp[u][1]=dp[u][0]=0;//1 切父节点变成链的最小花费，0不切
    int sum1 = 0;
    int maxn = 0, maxid = -1;
    int smaxn = 0, smaxid = -1;
    for(int i =head[u];~i;i=e[i].next)
    {
        int v = e[i].v;
        if(v == fa)continue;
        dfs(v,u);
        sum1 += dp[v][1]+2;
        int tmp = dp[v][0] - (dp[v][1] + 2);
        tmp = -tmp;
        if(tmp > smaxn)
        {
            smaxn = tmp;
            smaxid = v;
            if(smaxn > maxn)
            {
                swap(smaxn,maxn);
                swap(smaxid,maxid);
            }
        }
    }
    dp[u][0] = sum1 - maxn;
    dp[u][1] = sum1 - maxn - smaxn;
}
void find_bcc()
{ ans=0;
  dfs(1,-1);
  printf("%d\n",dp[1][1]+1);
}
int main()
{
  int cas;
  while(~scanf("%d",&cas))
  { int a,b;
    while(cas--)
    { data();
      scanf("%d",&n);
      FOR(i,2,n)
      {
        scanf("%d%d",&a,&b);add(a,b);add(b,a);
      }
      find_bcc();
    }

  }
}