【CF1095F】 Make It Connected（最小生成樹）

題目鏈接
如果沒有特殊邊的話顯然答案就是權值最小的點向其他所有點連邊。
所以把特殊邊和權值最小的點向其他點連的邊丟一起跑最小生成樹就行了。

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 200010;
typedef long long ll;
inline ll read(){
    ll s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
struct Edge{
    int from, to;
    ll dis;
    int operator < (const Edge A) const{
        return dis < A.dis;
    }
}e[MAXN << 1];
int n, m, pos;
ll ans, w[MAXN], mn = 2147483647214748364ll;
int f[MAXN];
inline int find(int x){
    return f[x] == x ? x : f[x] = find(f[x]);
}
int main(){
    n = read(); m = read();
    for(int i = 1; i <= n; ++i){
        w[f[i] = i] = read();
        if(w[i] < mn) mn = w[i], pos = i;
    }
    for(int i = 1; i <= m; ++i){
        e[i].from = read();
        e[i].to = read();
        e[i].dis = read();
    }
    for(int i = 1; i <= n; ++i)
        e[i + m] = (i == pos) ? (Edge){ 0, 0, 2147483647214748364ll } : (Edge){ i, pos, w[i] + w[pos] };
    sort(e + 1, e + n + m + 1);
    for(int i = 1, k = n - 1; k; ++i)
        if(find(e[i].from) != find(e[i].to))
            f[find(e[i].from)] = find(e[i].to), --k, ans += e[i].dis;
    printf("%lld\n", ans);
    return 0;
}