HDU 4714 Tree2cycle DP 2013杭电热身赛 1009

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4714

Tree2cycle

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 400    Accepted Submission(s): 78

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

 

题目大意：已知一棵树，切边与连边都需要1的cost。现在要把这棵树分割连接成为一个环，求最小cost。

 

解题思路：将整棵树拆成N个单枝(所有点的度小于2)，所需的cost即为2*N-1(拆成N个单枝需 N-1 cost，合成一个环需要 N cost)。

可用树形DP求出N最小的情况，用dp[i][j]表示以i为根的可用度为j的最小单枝数。

状态转移方程：dp[root][0]=min(dp[root][0]+dp[son][0],dp[root][1]+dp[son][1]-1)

　　　　　　　dp[root][1]=min(dp[root][1]+dp[son][0],dp[root][2]+dp[son][1]-1)

　　　　　　　dp[root][2]=dp[root][2]+dp[son][0]

P.S. 该题的数据量较大，要加#pragma comment(linker,"/STACK:1024000000,1024000000")语句，而且要用C++编译器，不能用G++。

 

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<iostream>

 4 #include<algorithm>

 5 #pragma comment(linker,"/STACK:1024000000,1024000000")

 6 using namespace std;

 7 #define N 1000005

 8 #define M 2000010

 9 int vis[N],all,first[N],next[M],v[M],e,degree[N],dp[N][3];

10 void addedge(int x,int y)

11 {

12     v[e]=y;

13     next[e]=first[x];

14     first[x]=e++;

15 }

16 void dfs(int root)

17 {

18     int i,k;

19     dp[root][0]=dp[root][1]=dp[root][2]=1;

20     vis[root]=1;

21     for(i=first[root];i!=-1;i=next[i])

22     {

23         k=v[i];

24         if(!vis[k])

25         {

26             dfs(k);

27             dp[root][0]=min(dp[root][0]+dp[k][0],dp[root][1]+dp[k][1]-1);

28             dp[root][1]=min(dp[root][1]+dp[k][0],dp[root][2]+dp[k][1]-1);

29             dp[root][2]=dp[root][2]+dp[k][0];

30         }

31     }

32 }

33 int main()

34 {

35     int t,i,x,y;

36     scanf("%d",&t);

37     while(t--)

38     {

39         memset(dp,0,sizeof(dp));

40         memset(first,-1,sizeof(first));

41         e=0;

42         all=0;

43         int n;

44         scanf("%d",&n);

45         for(i=1;i<n;i++)

46         {

47             scanf("%d%d",&x,&y);

48             addedge(x,y);

49             addedge(y,x);

50         }

51         memset(vis,0,sizeof(vis));

52         dfs(1);

53         all=min(dp[1][0],min(dp[1][1],dp[1][2]));

54         // cout<<all<<endl;

55         printf("%d\n",all*2-1);

56     }

57     return 0;

58 }

View Code

 

本人入ACM时间尚短，写的不好的地方请多见谅。