Link: D. Shichikuji and Power Grid

D. Shichikuji and Power Grid
Description: Shichikuji is the new resident deity of the South Black Snail Temple. Her first job is as follows:

There are n new cities located in Prefecture X. Cities are numbered from 1 to n. City i is located xi km North of the shrine and yi km East of the shrine. It is possible that (xi,yi)=(xj,yj) even when i≠j.

Shichikuji must provide electricity to each city either by building a power station in that city, or by making a connection between that city and another one that already has electricity. So the City has electricity if it has a power station in it or it is connected to a City which has electricity by a direct connection or via a chain of connections.

Building a power station in City i will cost ci yen;
Making a connection between City i and City j will cost ki+kj yen per km of wire used for the connection. However, wires can only go the cardinal directions (North, South, East, West). Wires can cross each other. Each wire must have both of its endpoints in some cities. If City i and City j are connected by a wire, the wire will go through any shortest path from City i to City j. Thus, the length of the wire if City i and City j are connected is |xi−xj|+|yi−yj| km.
Shichikuji wants to do this job spending as little money as possible, since according to her, there isn’t really anything else in the world other than money. However, she died when she was only in fifth grade so she is not smart enough for this. And thus, the new resident deity asks for your help.

And so, you have to provide Shichikuji with the following information: minimum amount of yen needed to provide electricity to all cities, the cities in which power stations will be built, and the connections to be made.

If there are multiple ways to choose the cities and the connections to obtain the construction of minimum price, then print any of them.

Input
First line of input contains a single integer n (1≤n≤2000) — the number of cities.

Then, n lines follow. The i-th line contains two space-separated integers xi (1≤xi≤1e6) and yi (1≤yi≤106) — the coordinates of the i-th city.

The next line contains n space-separated integers c1,c2,…,cn (1≤ci≤1e9) — the cost of building a power station in the i-th city.

The last line contains n space-separated integers k1,k2,…,kn (1≤ki≤1e9).

Output
In the first line print a single integer, denoting the minimum amount of yen needed.

Then, print an integer v — the number of power stations to be built.

Next, print v space-separated integers, denoting the indices of cities in which a power station will be built. Each number should be from 1 to n and all numbers should be pairwise distinct. You can print the numbers in arbitrary order.

After that, print an integer e — the number of connections to be made.

Finally, print e pairs of integers a and b (1≤a,b≤n, a≠b), denoting that a connection between City a and City b will be made. Each unordered pair of cities should be included at most once (for each (a,b) there should be no more (a,b) or (b,a) pairs). You can print the pairs in arbitrary order.

If there are multiple ways to choose the cities and the connections to obtain the construction of minimum price, then print any of them.

Examples
input
3
2 3
1 1
3 2
3 2 3
3 2 3
output
8
3
1 2 3
0
input
3
2 1
1 2
3 3
23 2 23
3 2 3
output
27
1
2
2
1 2
2 3
Note
For the answers given in the samples, refer to the following diagrams (cities with power stations are colored green, other cities are colored blue, and wires are colored red):

For the first example, the cost of building power stations in all cities is 3+2+3=8. It can be shown that no configuration costs less than 8 yen.

For the second example, the cost of building a power station in City 2 is 2. The cost of connecting City 1 and City 2 is 2⋅(3+2)=10. The cost of connecting City 2 and City 3 is 3⋅(2+3)=15. Thus the total cost is 2+10+15=27. It can be shown that no configuration costs less than 27 yen.

Problem solving:
这道题的意思就是给你n个城市,现在要为这n个城市通电。每个城市自己建造一座电站的花费告诉我们了,每两个城市通过电线连起来的花费是这两个城市的曼哈顿距离与这两个城市拉电线的花费之和的乘积。现在问你能给n个城市通电的最小花费。

我们可以用最小生成树写。首先设置一个超级源点,让这个点与n个城市建边,边的权值为这个城市建造电站的花费。再给每两个城市之间建边,边的权值是这两个城市拉电线的花费。然后最小生成树写一发就行了。还有个很恶心的东西就是输出的控制。不过也很好实现。

Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e6 + 10;
ll        f[maxn];
struct node
{
    ll x, y;
} p[maxn];
struct Node
{
    ll u, v, w;
} P[maxn];
bool cmp(Node a, Node b)
{
    return a.w < b.w;
}
#define pil    pair<ll, ll>
bool cmp2(pil a, pil b)
{
    return a.first < b.first;
}
ll find(ll x)
{
    return x != f[x] ? f[x] = find(f[x]) : x;
}
void join(ll x, ll y)
{
    x = find(x), y = find(y);
    if (x != y)
        f[x] = y;
}
vector<pil> Ans, AAns;
ll          r[maxn];
int main()
{
    ios::sync_with_stdio(0);
    ll co, n, pos;
    cin >> n; pos = n + 1;
    for (int i = 0; i <= n; i++)
        f[i] = i;
    for (int i = 1; i <= n; i++)
    {
        cin >> p[i].x >> p[i].y;
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> co; P[i].u = i, P[i].v = 0, P[i].w = co;
    }
    for (int i = 1; i <= n; i++)
        cin >> r[i];
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
        {
            ll mid = (abs(p[i].x - p[j].x) + abs(p[i].y - p[j].y)) * (r[i] + r[j]);
            P[pos].u = j, P[pos].v = i, P[pos++].w = mid;
        }
    sort(P + 1, P + 1 + pos, cmp);
    ll ans = 0, MMid = 0, mmid = 0;
    for (int i = 1; i <= pos; i++)
    {
        if (find(P[i].u) != find(P[i].v))
        {
            join(P[i].u, P[i].v), ans += P[i].w;
            if (P[i].u != 0 && P[i].v != 0)
                MMid++, Ans.push_back(make_pair(P[i].u, P[i].v));
            else
                mmid++, AAns.push_back(make_pair(P[i].u, P[i].v));
        }
    }
    cout << ans << endl;
    cout << mmid << endl;
    sort(AAns.begin(), AAns.end(), cmp2);
    for (int i = 0; i < AAns.size(); i++)
    {
        ll a1 = AAns[i].first, a2 = AAns[i].second;
        if (a1)
            cout << a1 << " ";
        else
            cout << a2 << " ";
    }
    cout << endl;
    cout << MMid << endl;
    for (int i = 0; i < Ans.size(); i++)
    {
        ll a1 = Ans[i].first, a2 = Ans[i].second;
        if (a1 > a2)
            swap(a1, a2);
        cout << a1 << " " << a2 << endl;
    }
}

可能我的代码写得有点丑,不过思想到位了就行。我也懒得改了。