『Balancing Act 树的重心』

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树的重心

我们先来认识一下树的重心。

树的重心也叫树的质心。找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后，生成的多棵树尽可能平衡。

根据树的重心的定义，我们可以通过树形DP来求解树的重心。

设\(Max_i\)代表删去i节点后树中剩下子树中节点最多的一个子树的节点数。由于删去节点i至少将原树分为两部分，所以满足\(\ \frac{1}{2} \leq Max_i\)，我们要求的就是一个\(i\)，使得\(Max_i\)最小。

对于Max数组，我们可以列出如下状态转移方程:
\[ Max_i=\max_{y\in son(i)}\{size_y,n-size_x\} \]

size数组即为节点个数(树的大小)，可以在树形DP中顺带求解。

\(Code:\)

inline void dp(int r,int f)
{
    size[r]=1;
    for(int i=0;i<Link[r].size();i++)
    {
        int Son=Link[r][i];
        if(Son==f)continue;
        dp(Son,r);
        size[r]+=size[Son];
        Max[r]=max(Max[r],size[Son]);
    }
    Max[r]=max(Max[r],n-size[r]);
    if(Max[r]==Max[ans]&&r<ans)ans=r;
    if(Max[r]<Max[ans])ans=r;
}

还是通过一道例题来认识一下。

Balancing Act(POJ1655)

Description

The city consists of intersections and streets that connect them.

Heavy snow covered the city so the mayor Milan gave to the winter-service a list of streets that have to be cleaned of snow. These streets are chosen such that the number of streets is as small as possible but still every two intersections to be connected i.e. between every two intersections there will be exactly one path. The winter service consists of two snow plovers and two drivers, Mirko and Slavko, and their starting position is on one of the intersections.

The snow plover burns one liter of fuel per meter (even if it is driving through a street that has already been cleared of snow) and it has to clean all streets from the list in such order so the total fuel spent is minimal. When all the streets are cleared of snow, the snow plovers are parked on the last intersection they visited. Mirko and Slavko don’t have to finish their plowing on the same intersection.

Write a program that calculates the total amount of fuel that the snow plovers will spend.

Input Format

The first line of the input contains two integers: N and S, 1 <= N <= 100000, 1 <= S <= N. N is the total number of intersections; S is ordinal number of the snow plovers starting intersection. Intersections are marked with numbers 1...N.

Each of the next N-1 lines contains three integers: A, B and C, meaning that intersections A and B are directly connected by a street and that street‘s length is C meters, 1 <= C <= 1000.

Output Format

Write to the output the minimal amount of fuel needed to clean all streets.

Sample Input

5 2
1 2 1
2 3 2
3 4 2
4 5 1

Sample Output

6

解析

这个就是树的重心的模板题了嘛。

还有题目的第二问就是重心子树中节点数最多的子树的节点数。嗯！刚好符合我们Max数组的定义，直接输出就可以了。

\(Code:\)

#include<cstdio>
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#define mset(name,val) memset(name,val,sizeof name)
using namespace std;
const int N=100000+50;
int n,s,sum,vis[N],dis[N];
struct edge{int val,ver;};
vector < edge > Link[N];
inline void input(void)
{
    scanf("%d%d",&n,&s);
    for(int i=1;i<n;i++)
    {
        int x,y,v;
        scanf("%d%d%d",&x,&y,&v);
        Link[x].push_back((edge){v,y});
        Link[y].push_back((edge){v,x});
        sum+=v*2;
    }
}
inline int Search(int start)
{
    queue< int >q;
    mset(vis,0x00);
    mset(dis,0x00);
    vis[start]=1;
    q.push(start);
    while(!q.empty())
    {
        int temp=q.front();q.pop();
        for(int i=0;i<Link[temp].size();i++)
        {
            if(!vis[Link[temp][i].ver])
            {
                vis[Link[temp][i].ver]=true;
                dis[Link[temp][i].ver]=dis[temp]+Link[temp][i].val;
                q.push(Link[temp][i].ver);
            }
        }
    }
    int res=0,Maxdis=0;
    for(int i=1;i<=n;i++)
    {
        if(dis[i]>Maxdis)
        {
            Maxdis=dis[i];
            res=i;
        }
    }
    return res;
}
int main(void)
{
    input();
    int p=Search(s);
    printf("%d\n",sum-dis[Search(p)]);
    return 0;
}

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