题目链接:http://codeforces.com/problemset/problem/437/D
思路:并差集应用,先对所有的边从大到小排序,然后枚举边的时候,如果某条边的两个顶点不在同一个集合中就合并,并且用一个sum记录这两个集合的大小,这样这两个集合中的每一对点都要经过这条边,然后更新一下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 ", 2.0 * ans / (1.0 * N * (N - 1))); return 0; }