旅行

链接：https://ac.nowcoder.com/acm/contest/7329/C
题解：
在选择排列的时候，从边权最大的递减构造，那么这条边就是当前待选择边权中的最大值了。因此构造一个最大生成树即可。

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define SZ(x) ((int)(x).size())
#define pb push_back
#define pii pair<int, int>
#define mset(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int N = 5e5+10;
const double eps = 1e-11;

int n, m, u[N], v[N], w[N], r[N], p[N];
bool cmp(const int i, const int j){return w[i] > w[j];}
int find(int x){return p[x] == x ? x : p[x] = find(p[x]);}
ll Kruskal(){
    ll ans = 0;
    for(int i=1; i<=n; i++) p[i] = i;
    for(int i=1; i<=m; i++) r[i] = i;
    sort(r+1, r+m+1, cmp);
    for(int i=1; i<=m; i++){
        int e = r[i]; int x = find(u[e]); int y = find(v[e]);
        if(x != y) {
            ans += w[e]; p[x] = y;
        }
    }
    return ans;
}
int main()
{
    n = read(), m = read();
    for(int i=1; i<=m; i++){
        u[i] = read(), v[i] = read(), w[i] = read();
    }
    print(Kruskal());

}