(思维+最小生成树)Codeforces1245D.Shichikuji and Power Grid

题意：(来源洛谷)

 

 

思路：一个非常巧妙的方法，将在城市i建造发电站看作是将城市i与城市0连接，代价即为ci。其他城市间则按题意建立边，求一次最小生成树即可。需要开long long。

代码：

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 2020;
struct edge{
    ll u, v, w;
    edge(ll a, ll b, ll c): u(a), v(b), w(c) {}
    bool operator < (const edge &x) const {
        return w < x.w;
    }
};
ll k[N], cost[N], x[N], y[N], f[N];
ll n;
vector <edge> e;
ll find(ll x) {
    return x == f[x] ? x : f[x] = find(f[x]);
}

int main()
{
    scanf("%lld", &n);
    for (ll i = 1; i <= n; i++)
        scanf("%lld %lld", &x[i], &y[i]);
    for (ll i = 1; i <= n; i++) scanf("%lld", &cost[i]);
    for (ll i = 1; i <= n; i++) scanf("%lld", &k[i]);
    
    for (ll i = 1; i <= n; i++) e.push_back(edge(0, i, cost[i]));
    for (ll i = 1; i <= n; i++)
        for (ll j = 1; j <= n; j++) {
            if (j <= i) continue;
            e.push_back(edge(i, j, (k[i]+k[j])*(abs(x[i]-x[j])+abs(y[i]-y[j]))));
        }
    sort(e.begin(), e.end());
    vector <ll> a;
    vector <edge> used;
    ll cnt = 0, ans = 0;
    for (ll i = 0; i <= n; i++) f[i] = i;
    for (auto it : e) {
        ll u = find(it.u), v = find(it.v);
        if (u != v) {
            ll z = 0;
            if (z == it.v) a.push_back(it.u);
            if (z == it.u) a.push_back(it.v);
            if (z != it.u && z != it.v) used.push_back(it);
            cnt++; f[u] = v; ans += it.w;
        }
        if (cnt == n) break;
    }
    cout << ans << endl;
    cout << a.size() << endl;
    for (auto it : a) printf("%lld ", it);
    puts("");
    cout << used.size() << endl;
    for (auto it : used) printf("%lld %lld
", it.u, it.v);
    return 0;
}