洛谷 P2986 [USACO10MAR]Great Cow Gat…(树形dp+容斥原理)

P2986 [USACO10MAR]伟大的奶牛聚集Great Cow Gat…

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。

输入输出格式

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

输入输出样例

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：

15 

/*
树形dp+容斥原理
f[i]表示i为关键点答案
dis[i]表示以i为根子树和,第一遍树形dp统计
num[i]表示点权和,tot为总点权和
容斥原理: f[v]=f[now]+(tot-num[v])*e[i].d-num[v]*e[i].d; 第二遍树形dp统计
由父亲节点转移到儿子节点 容斥比较难想一些,建议画图比照代码理解
一定明确要求Σ点权*边权最小，容斥的时候注意计算的是那个点的点权*哪条边的边权
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100007

using namespace std;
int head[N],w[N];
long long f[N],num[N],dis[N],tot;
int n,m,x,y,z,cnt;
struct edge
{
    int u,to,pre,d;
}e[N<<];

inline void add(int u,int to,int d)
{
    e[++cnt].to=to;e[cnt].d=d;e[cnt].pre=head[u];head[u]=cnt;
}

void tree_dp1(int fa,int now)
{
    num[now]=w[now];
    for(int i=head[now];i;i=e[i].pre)
    {
        if(e[i].to==fa) continue;
        tree_dp1(now,e[i].to);
        num[now]+=num[e[i].to];
        dis[now]+=dis[e[i].to]+num[e[i].to]*e[i].d;
    }
}

long long mn=999999999999999LL;

void tree_dp2(int fa,int now)
{
    mn=min(mn,f[now]);
    for(int i=head[now];i;i=e[i].pre)
    {
        int v=e[i].to;
        if(v==fa) continue;
        f[v]=f[now]+(tot-num[v])*e[i].d-num[v]*e[i].d;
        tree_dp2(now,v);
    }
}

int main()
{
    scanf("%d",&n);
    for(int i=;i<=n;i++)
      scanf("%d",&w[i]),tot+=w[i];
    for(int i=;i<n;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);add(y,x,z);
    }
    tree_dp1(,);
    f[]=dis[];
    tree_dp2(,);
    printf("%lld\n",mn);
    return ;
}