Codeforces Round #548 (Div. 2) C. Edgy Trees（并查集+快速幂）

思路用并查集统计一个连通块的节点个数，最后用总的减去他，设x是连通块的节点个数，o个联通块

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define ll long long
#define ld long double
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 200010
#define mod 1000000007
using namespace std;
int n, ans, k, f, mp[Max], u, sz[Max];
int F(int x) {
    return x == mp[x] ? x : mp[x] = F(mp[x]);
}
int poww(int a, int b) {
    ll ans = 1, base = a;
    while (b != 0) {
        if (b & 1 != 0)
            ans *= base, ans %= mod;
        base *= base, base %= mod;
        b >>= 1;
    }
    return ans % mod;
}
int main() {
    cin >> n >> k;
    for(int i = 1; i <= n; i++)
        mp[i] = i, sz[i] = 1;
    for(int a, b, c, i = 1; i < n; i++) {
        cin >> a >> b >> c;
        if(!c) {
            int x = F(a), y = F(b);
            mp[x] = y, sz[y] += sz[x];
        }
    }
    for(int i = 0; i <= n; i++)
        if(mp[i] == i)
            (ans += poww(sz[i], k)) %= mod;
    printf("%d", (poww(n, k) - ans + mod) % mod);
    return 0;
}