【洛谷2986】【USACO10MAR】伟大的奶牛聚集

题面

题目描述

Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place.

Each cow lives in one of N (1 <= N <= 100,000) different barns (conveniently numbered 1..N) which are connected by N-1 roads in such a way that it is possible to get from any barn to any other barn via the roads. Road i connects barns A_i and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has length L_i (1 <= L_i <= 1,000). The Great Cow Gathering can be held at any one of these N barns. Moreover, barn i has C_i (0 <= C_i <= 1,000) cows living in it.

When choosing the barn in which to hold the Cow Gathering, Bessie wishes to maximize the convenience (which is to say minimize the inconvenience) of the chosen location. The inconvenience of choosing barn X for the gathering is the sum of the distances all of the cows need to travel to reach barn X (i.e., if the distance from barn i to barn X is 20, then the travel distance is C_i*20). Help Bessie choose the most convenient location for the Great Cow

Gathering.

Consider a country with five barns with [various capacities] connected by various roads of varying lengths. In this set of barns, neither barn 3 nor barn 4 houses any cows.

1 3 4 5

@--1--@--3--@--3--@[2]

[1] |

2 | @[1] 2 Bessie can hold the Gathering in any of five barns; here is the table of inconveniences calculated for each possible location:

Gather ----- Inconvenience ------

Location B1 B2 B3 B4 B5 Total

1 0 3 0 0 14 17

2 3 0 0 0 16 19

3 1 2 0 0 12 15

4 4 5 0 0 6 15

5 7 8 0 0 0 15

If Bessie holds the gathering in barn 1, then the inconveniences from each barn are:

Barn 1 0 -- no travel time there!

Barn 2 3 -- total travel distance is 2+1=3 x 1 cow = 3 Barn 3 0 -- no cows there!

Barn 4 0 -- no cows there!

Barn 5 14 -- total travel distance is 3+3+1=7 x 2 cows = 14 So the total inconvenience is 17.

The best possible convenience is 15, achievable at by holding the Gathering at barns 3, 4, or 5.

Bessie正在计划一年一度的奶牛大集会，来自全国各地的奶牛将来参加这一次集会。当然，她会选择最方便的地点来举办这次集会。

每个奶牛居住在 N(1<=N<=100,000) 个农场中的一个，这些农场由N-1条道路连接，并且从任意一个农场都能够到达另外一个农场。道路i连接农场A_i和B_i(1 <= A_i <=N; 1 <= B_i <= N),长度为L_i(1 <= L_i <= 1,000)。集会可以在N个农场中的任意一个举行。另外，每个牛棚中居住者C_i(0 <= C_i <= 1,000)只奶牛。

在选择集会的地点的时候，Bessie希望最大化方便的程度(也就是最小化不方便程度)。比如选择第X个农场作为集会地点，它的不方便程度是其它牛棚中每只奶牛去参加集会所走的路程之和，(比如，农场i到达农场X的距离是20，那么总路程就是C_i*20)。帮助Bessie找出最方便的地点来举行大集会。

输入格式：

Line 1: A single integer: N

Lines 2..N+1: Line i+1 contains a single integer: C_i

Lines N+2..2*N: Line i+N+1 contains three integers: A_i, B_i, and L_i

输出格式：

Line 1: The minimum inconvenience possible

输入样例#1：

5

1

1

0

0

2

1 3 1

2 3 2

3 4 3

4 5 3

输出样例#1：

题解

考虑如果依次枚举每一个点作为集会的地点

使用DFS进行计算

然后再依次比较

时间复杂度O(n^2)

但是n的范围太大，显然会超时。

那么，我们应当如何优化？

先看看样例

通过一次O(n)的计算，很容易得出来

如果选择1号节点，答案就是17

既然O(n^2)的计算无法在时间内求解

那么是否可以递推出来呢？

显然是可以的。

观察如果已经知道1号节点所需的时间

那么，我们可以做如下假设：

① 所有的牛首先到达了1号节点

② 3号节点以及他子树上的节点都需要退回1->3的路径的长度

③ 除了3号节点以及他子树上的节点都需要前进1->3的路径的长度

通过上面的三条东西，我们就可以从任意一个父节点推出子节点的时间

所以，又是一遍O(n)的计算就可以推出最终的答案

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define MAX 200100
#define ll long long
inline ll read()
{
      register ll x=0,t=1;
      register char ch=getchar();
      while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
      if(ch=='-'){t=-1;ch=getchar();}
      while(ch<='9'&&ch>='0'){x=x*10+ch-48;ch=getchar();}
      return x*t;
}
ll dis[MAX],C[MAX],Q[MAX],f[MAX],Sum,Ans=1000000000000000000;
struct Line
{
      ll v,next,w;
}e[MAX];
ll h[MAX],cnt=1,N;
inline void Add(ll u,ll v,ll w)
{
      e[cnt]=(Line){v,h[u],w};
      h[u]=cnt++;
}
//使用两遍DFS
//第一遍以任意点为根节点计算一遍
//dis[i]表示以i为根的子树到根的距离之和
ll DFS(ll u,ll ff)
{
      ll tot=0;
      for(ll i=h[u];i;i=e[i].next)
      {
               ll v=e[i].v;
               if(v!=ff)
               {
                      ll s=DFS(v,u);//子树上牛的数量
                      dis[u]+=dis[v]+e[i].w*s;//统计
                   tot+=s;//牛的个数
               }
      }
      return Q[u]=tot+C[u];
}
//第二遍计算偏移后的值
//先可以假设走到当前节点的父节点
//再让当前自己点所有牛退回来，父节点的所有牛走过去即可
void DFS2(ll u,ll ff)
{
       for(ll i=h[u];i;i=e[i].next)
       {
                  ll v=e[i].v;
                  if(v!=ff)
                  {
                           ll ss=e[i].w;
                           f[v]=f[u]-Q[v]*ss+(Sum-Q[v])*ss;
                           DFS2(v,u);
                  }
       }
}
int main()
{
      N=read();
      for(ll i=1;i<=N;++i)
        C[i]=read();
      for(ll i=1;i<=N;++i)
        Sum+=C[i];//统计牛的总数
      for(ll i=1;i<N;++i)
      {
                 ll u=read(),v=read(),w=read();
                 Add(u,v,w);
                 Add(v,u,w);
      }
      DFS(1,1);//求出以1为聚集处的结果 
      DFS2(1,1);//求出其他的偏移值
      for(ll i=1;i<=N;++i)
            Ans=min(Ans,f[i]);
        cout<<Ans+dis[1]<<endl;
        return 0;
}