Codeforces Round#250 D. The Child and Zoo(並差集)
阿新 • • 發佈:2018-12-24
思路:並差集應用,先對所有的邊從大到小排序,然後列舉邊的時候,如果某條邊的兩個頂點不在同一個集合中就合併,並且用一個sum記錄這兩個集合的大小,這樣這兩個集合中的每一對點都要經過這條邊,然後更新一下sum就可以了。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #define REP(i, a, b) for (int i = (a); i < (b); ++i) #define FOR(i, a, b) for (int i = (a); i <= (b); ++i) using namespace std; const int MAX_N = (100000 + 100); struct Edge { int u, v, w; } edge[MAX_N << 1]; int cmp(const Edge &p, const Edge &q) { return p.w > q.w; } int N, M, a[MAX_N], parent[MAX_N]; long long sum[MAX_N], ans; int Find(int x) { if (x == parent[x]) return x; return parent[x] = Find(parent[x]); } int main() { cin >> N >> M; FOR(i, 1, N) cin >> a[i], parent[i] = i, sum[i] = 1; FOR(i, 1, M) { int u, v; cin >> u >> v; edge[i].u = u, edge[i].v = v, edge[i].w = min(a[u], a[v]); } sort(edge + 1, edge + M + 1, cmp); ans = 0; FOR(i, 1, M) { int r1 = Find(edge[i].u), r2 = Find(edge[i].v); if (r1 == r2) continue; parent[r1] = r2; ans += edge[i].w * sum[r1] * sum[r2]; sum[r2] += sum[r1]; } printf("%.7f\n", 2.0 * ans / (1.0 * N * (N - 1))); return 0; }