HDU 4714 Tree2cycle （树形DP）

Tree2cycle

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 324    Accepted Submission(s): 54


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
 

 

 

简单树形DP来一发

 

用dp1表示形成一颗链，而且该点在端点。

dp2表示形成一颗链需要的最少步数

 

 1 /* *******************************************
 2 Author       : kuangbin
 3 Created Time : 2013年09月08日 星期日 12时00分01秒
 4 File Name    : 1009.cpp
 5 ******************************************* */
 6 #pragma comment(linker, "/STACK:1024000000,1024000000")
 7 #include <stdio.h>
 8 #include <algorithm>
 9 #include <iostream>
10 #include <string.h>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 
21 const int MAXN = 1000010;
22 vector<int>vec[MAXN];
23 int dp1[MAXN];
24 int dp2[MAXN];
25 void dfs(int u,int pre)
26 {
27     int sz = vec[u].size();
28     int sum1 = 0;
29     int maxn = 0, maxid = -1;
30     int smaxn = 0, smaxid = -1;
31     for(int i = 0;i < sz;i++)
32     {
33         int v = vec[u][i];
34         if(v == pre)continue;
35         dfs(v,u);
36         sum1 += dp2[v]+2;
37         int tmp = dp1[v] - (dp2[v] + 2);
38         tmp = -tmp;
39         if(tmp > smaxn)
40         {
41             smaxn = tmp;
42             smaxid = v;
43             if(smaxn > maxn)
44             {
45                 swap(smaxn,maxn);
46                 swap(smaxid,maxid);
47             }
48         }
49     }
50     dp1[u] = sum1 - maxn;
51     dp2[u] = sum1 - maxn - smaxn;
52 
53 }
54 int main()
55 {
56     //freopen("in.txt","r",stdin);
57     //freopen("out.txt","w",stdout);
58     int T;
59     int n;
60     scanf("%d",&T);
61     while(T--)
62     {
63         scanf("%d",&n);
64         int u,v;
65         for(int i = 1;i <= n;i++)
66             vec[i].clear();
67         for(int i = 1;i < n;i++)
68         {
69             scanf("%d%d",&u,&v);
70             vec[u].push_back(v);
71             vec[v].push_back(u);
72         }
73         dfs(1,-1);
74         cout<<dp2[1]+1<<endl;
75     }
76     return 0;
77 }

 

 

 

 

posted on 2013-09-08 19:51  kuangbin  阅读(1260)  评论(0)  编辑  收藏  举报

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