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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, please output one integer representing minimal cost to transform the tree to a cycle.

[align=left]Sample Input[/align]

1
4
1 2
2 3
2 4

[align=left]Sample Output[/align]

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.

[align=left]Source[/align]
2013 ACM/ICPC Asia Regional Online —— Warmup

[align=left]Recommend[/align]
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

/* *******************************************
Author       : kuangbin
Created Time : 2013年09月08日 星期日 12时00分01秒
File Name    : 1009.cpp
******************************************* */
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 1000010;
vector<int>vec[MAXN];
int dp1[MAXN];
int dp2[MAXN];
void dfs(int u,int pre)
{
int sz = vec[u].size();
int sum1 = 0;
int maxn = 0, maxid = -1;
int smaxn = 0, smaxid = -1;
for(int i = 0;i < sz;i++)
{
int v = vec[u][i];
if(v == pre)continue;
dfs(v,u);
sum1 += dp2[v]+2;
int tmp = dp1[v] - (dp2[v] + 2);
tmp = -tmp;
if(tmp > smaxn)
{
smaxn = tmp;
smaxid = v;
if(smaxn > maxn)
{
swap(smaxn,maxn);
swap(smaxid,maxid);
}
}
}
dp1[u] = sum1 - maxn;
dp2[u] = sum1 - maxn - smaxn;

}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int u,v;
for(int i = 1;i <= n;i++)
vec[i].clear();
for(int i = 1;i < n;i++)
{
scanf("%d%d",&u,&v);
vec[u].push_back(v);
vec[v].push_back(u);
}
dfs(1,-1);
cout<<dp2[1]+1<<endl;
}
return 0;
}