Codeforces Round #529 (Div. 3) F. Make It Connected（最小生成树）

AC

最小生成树，建边的时候不需要N2{N^2}N2，首先N个点需要N-1条边，N个点的权值都会被计算，一些顶点会重复计算。我们让重复计算的点是权值最小的点，就可以保证生成树权值最小
将特殊边加到边集里，跑一边Kruskal

#include <bits/stdc++.h>
#define P pair<int, int>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define REP(i, n) for (int i = 1; i <= (n); ++i)
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define N 200006
#define LL  long long
using namespace std;
struct ac{LL u, v, d;
}g[N*2];
LL a[N];
int fa[N];
int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);
}
int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifint n, m;while (scanf("%d %d", &n, &m) != EOF) {int pos = 1;REP (i, n) {scanf("%lld", &a[i]);if (a[i] < a[pos])  pos = i;}int len = 0;REP (i, n) {if (i == pos)   continue;g[len].u = pos;g[len].v = i;g[len].d = a[pos] + a[i];len++;}rep (i, m) {scanf("%lld %lld %lld", &g[len].u, &g[len].v, &g[len].d);++len;}sort(g, g + len, [&](const ac &x, const ac &y){return x.d < y.d;});REP (i, n)  fa[i] = i;LL u, v, d, ans = 0, cnt = 0;rep (i, len) {u = g[i].u;v = g[i].v;d = g[i].d;if (find(u) != find(v)) {fa[find(u)] = find(v);ans += d;if (++cnt == n-1)   break;}}printf("%lld\n", ans);}return 0;
}

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Codeforces Round #529 (Div. 3) F. Make It Connected（最小生成树）

AC

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