Codeforces Round #548 (Div. 2) C. Edgy Trees（dfs || 并查集)

题目链接：https://codeforces.com/contest/1139/problem/C

题意：给了一棵树，n个点，m条边。让从中选k个点，使得从a1到a2，a2到a3，ak-1到ak的路径中至少经过一条黑色的边，问这样的集合有多少个。

思路：在长度为k的序列里，由于每个位置有n种可能，所以总的可能数为 $n^{k}$ ，用总的可能数减去仅经过红边的序列数即可。然后统计只经过红边的节点在同一个连通区域的结果个数，减去输出即可。并查集另开一个c数组存节点个数。

dfs代码：

#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define IOS cin.tie(0); ios::sync_with_stdio(0);
const int mod = 1e9+7;
const int N = 1e6+7;
int ant, vis[N];
std::vector< int > v[N];
int dfs(int x)
{vis[x] = 1; ant++;for(int i = 0; i < v[x].size(); i++)if(!vis[v[x][i]]) dfs(v[x][i]);
}
ll poww(ll a, ll b)
{ll ans = 1;while (b){if(b & 1) ans = (ans * a) % mod;a = (a * a) % mod;b >>= 1;}return ans;
}
int main()
{IOS; int n, k;std::cin >> n >> k;for(int i = 1; i < n; i++){int a, b, x;std::cin >> a >> b >> x;if(!x){v[a].push_back(b);v[b].push_back(a);}}ll ans = poww(n, k), cnt = 0;for(int i = 1; i <= n; i++){ant = 0;if(!vis[i])dfs(i), (cnt += poww(ant, k)) %= mod;}std::cout << (ans - cnt + mod) % mod << '\n';
}

并查集代码：

#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define IOS cin.tie(0); ios::sync_with_stdio(0);
const int mod = 1e9+7;
const int N = 1e6+7;
int f[N], c[N];
int findd(int x)
{if(x != f[x]) return f[x] = findd(f[x]);return x;
}
void join(int x, int y)
{int tx = findd(x), ty = findd(y);if(tx != ty) f[tx] = ty, c[ty] += c[tx];
}
ll poww(ll a, ll b)
{ll ans = 1;while (b){if(b & 1) ans = (ans * a) % mod;a = (a * a) % mod;b >>= 1;}return ans;
}
int main()
{IOS; int n, k;std::cin >> n >> k;for(int i = 1; i <= n; i++)f[i] = i, c[i] = 1;for(int i = 1; i < n; i++){int a, b, x;std::cin >> a >> b >> x;if(!x) join(a, b);}ll ans = poww(n, k), cnt = 0;for(int i = 1; i <= n; i++){if(findd(i) == i)(cnt += poww(c[i], k) ) %= mod;}std::cout << (ans - cnt + mod) % mod << '\n';
}

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