CF1095F Make It Connected

CF1095F Make It Connected

一眼kruskal板子题，又看了一眼被数据范围劝退，去原题看了一眼div3的题，仔细思考了一下。

我们把所有边考虑一下，除了特殊边，剩下边都是 (a_i + a_j) 的形式。那么显然，在固定一个的情况下，一个最小肯定最优，那么我们容易想到一定存在一个 (a_{mini}) 使得 (a_{i} + a_{mini} le a_{i} + a_{j}) 那么我们只需要 (a_{i}+a_{mini}) 的 (N-1) 条边和 (M) 条特殊边就行了。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

using namespace std;

typedef long long ll;
const ll MAXN = 1e6+10, INF = 0x3f3f3f3f3f3f3f3f;

ll N, M, fa[MAXN], val[MAXN];

struct edge {
    ll x, y, v;
    edge(ll _x, ll _y, ll _v): x(_x), y(_y), v(_v) {}
	edge() {} 
	friend bool operator < (edge a, edge b) {
        return a.v < b.v;
    }
} E[MAXN];

ll ans = 0, minn = INF, mini = 0;

ll find_(ll);
void onion(ll, ll);

int main() {
    scanf("%lld%lld", &N, &M);
    for (ll i = 1; i <= N; i++) { 
        scanf("%lld", val+i), fa[i] = i;
        if (minn > val[i])
        	minn = val[i], mini = i;
    }
    for (ll i = 1; i <= M; i++) { 
		scanf("%lld%lld%lld", &E[i].x, &E[i].y, &E[i].v);
		E[i].v = min(E[i].v, val[E[i].x] + val[E[i].y]);
	} 
	for (ll i = 1; i <= N; i++) if (i != mini)
		E[++M] = edge(i, mini, val[i] + val[mini]);
    sort(E+1, E+M+1);
    for (ll i = 1, qq = 1; i < N && qq <= M; i++, qq++) {
    	ll fx = find_(E[qq].x), fy = find_(E[qq].y);
		if (fx == fy) {i--; continue;}
		onion(fx, fy);
		ans += E[qq].v;
    }
    printf("%lld
", ans);
    return 0;
}

void onion(ll x, ll y) {
    ll fx = find_(x), fy = find_(y);
    if (fx != fy) {
        fa[fx] = fy;
        val[fy] = min(val[fy], val[fx]);
    }
}

ll find_(ll x) {return fa[x] == x ? x : fa[x] = find_(fa[x]);}