HDU 4118 Holiday's Accommodation(树形DP)

Holiday's Accommodation

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 200000/200000 K (Java/Others)
Total Submission(s): 2925    Accepted Submission(s): 894

Problem Description

Nowadays, people have many ways to save money on accommodation when they are on vacation.
One of these ways is exchanging houses with other people.
Here is a group of N people who want to travel around the world. They live in different cities, so they can travel to some other people's city and use someone's house temporary. Now they want to make a plan that choose a destination for each person. There are 2 rules should be satisfied:
1. All the people should go to one of the other people's city.
2. Two of them never go to the same city, because they are not willing to share a house.
They want to maximize the sum of all people's travel distance. The travel distance of a person is the distance between the city he lives in and the city he travels to. These N cities have N - 1 highways connecting them. The travelers always choose the shortest path when traveling.
Given the highways' information, it is your job to find the best plan, that maximum the total travel distance of all people.

 

 

Input

The first line of input contains one integer T(1 <= T <= 10), indicating the number of test cases.
Each test case contains several lines.
The first line contains an integer N(2 <= N <= 10⁵), representing the number of cities.
Then the followingN-1 lines each contains three integersX, Y,Z(1 <= X, Y <= N, 1 <= Z <= 10⁶), means that there is a highway between city X and city Y , and length of that highway.
You can assume all the cities are connected and the highways are bi-directional.

 

 

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y represents the largest total travel distance of all people.

 

 

Sample Input

2 4 1 2 3 2 3 2 4 3 2 6 1 2 3 2 3 4 2 4 1 4 5 8 5 6 5

 

 

Sample Output

Case #1: 18 Case #2: 62

 

 

Source

2011 Asia ChengDu Regional Contest

 

 

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/*
题意：n个结点，每个节点都有一个房子和一个人，这些人都想环游世界，也就是所有结点都去一遍，他们旅游的时候会选则最近的路线，
让你求所有人都达成愿望之后最多的路程

思路：这个题光是题意就看了很长时间，刚开始理解成一个人只要到达一个城市就好了，显然理解偏了。
每一条路最多能给总路程提供的价值：很显然就是这条路两边的相对较小的定点数×这条路的长的，然后dfs搜一遍就好了
*/
#include<stdio.h>
#include<vector>
#include<string.h>
#include<iostream>
#define N 100005
using namespace std;
struct node
{
    int to,len;//下一个节点是哪里，以to为重点的边多长
    node(int x=0,int y=0)
    {
        to=x;len=y;
    }
}fr[N];
vector<node > v[N];//构图用的数组
int t,n;
long long sum;
long long dp[N];//表示到i点位置，左边子树有多少个点
bool visit[N];//记录这个点走没走
void dfs(int id,int len)
{
    //cout<<"id="<<id<<endl;
    visit[id]=true;
    //cout<<dp[id]<<endl;
    for(int i=0;i<v[id].size();i++)
    {
        int next=v[id][i].to;//下一步要走的路；
        int next_len=v[id][i].len;//下一条路的长度
        if(visit[next]) continue;//这一步走了就跳过
        dfs(next,next_len);
        dp[id]+=dp[next];
    }
    dp[id]++;
    //cout<<"dp["<<id<<"]="<<dp[id]<<endl;
    sum+=(long long)min(dp[id],n-dp[id])*len;
}
int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int l=1;l<=t;l++)
    {
        sum=0;
        memset(dp,0,sizeof dp);
        memset(visit,false,sizeof visit);
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            v[i].clear();
        for(int i=1;i<n;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            //cout<<a<<" "<<b<<" "<<c<<endl;
            v[a].push_back(node(b,c));
            v[b].push_back(node(a,c));
        }    
        dfs(1,0);
        printf("Case #%d: %lld\n",l,sum*2);
    }
    return 0;
}