AC

• 最小生成樹，建邊的時(shí)候不需要$N^2$，首先N個(gè)點(diǎn)需要N-1條邊，N個(gè)點(diǎn)的權(quán)值都會(huì)被計(jì)算，一些頂點(diǎn)會(huì)重復(fù)計(jì)算。我們讓重復(fù)計(jì)算的點(diǎn)是權(quán)值最小的點(diǎn)，就可以保證生成樹權(quán)值最小
• 將特殊邊加到邊集里，跑一邊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; }

總結(jié)

以上是生活随笔為你收集整理的Codeforces Round #529 (Div. 3) F. Make It Connected（最小生成树）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。