[HDOJ5723]Abandoned country（最小生成树，期望）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5723

题意：求最小生成树，并且求这棵最小生成树上所有边走过次数的期望。

走过次数的期望=Σ边被走过次数*边权/(n*(n-1)/2)

求边被走过的次数，相当于关心这个边的两侧分别有多少点，走的次数就是两边的点数的乘积这个很好理解。可以从边的一侧开始dfs，找到这个边的一侧点数x，由于最小生成树是联通的，那么边的另一侧点数一定是n-x。所以这条边的贡献是(n-x)*x*w。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef struct Edge {
  int v, next;
  LL w;
}Edge;

typedef struct E {
  int u, v;
  LL w;
}E;
const int maxn = 200100;
const int maxm = 1000100;
int n, m;
E e[maxm<<1];
Edge edge[maxm<<1];
int head[maxn], ecnt;
int pre[maxn];
LL ret1, ret2;

int find(int x) {
  return x == pre[x] ? x : pre[x] = find(pre[x]);
}

bool unite(int x, int y) {
  x = find(x); y = find(y);
  if(x != y) {
    pre[x] = y;
    return 1;
  }
  return 0;
}

void init() {
  memset(head, -1, sizeof(head));
  ecnt = 0; ret1 = 0; ret2 = 0;
  for(int i = 1; i <= n; i++) pre[i] = i;
}

void adde(int u, int v, LL w) {
  edge[ecnt].v = v; edge[ecnt].w = w; edge[ecnt].next = head[u]; head[u] = ecnt++;
  edge[ecnt].v = u; edge[ecnt].w = w; edge[ecnt].next = head[v]; head[v] = ecnt++;
}

bool cmp(E a, E b) {
  return a.w < b.w;
}

int dfs(int u, int p) {
  int siz = 1;
  for(int i = head[u]; ~i; i=edge[i].next) {
    int v = edge[i].v; LL w = edge[i].w;
    if(p == v) continue;
    int pre = dfs(v, u);
    siz += pre;
    ret2 += (LL)pre * (LL)(n - pre) * w;
  }
  return siz;
}

int main() {
  //freopen("in", "r", stdin);
  int T, u, v;
  LL w;
  scanf("%d", &T);
  while(T--) {
    scanf("%d%d",&n,&m);
    init();
    for(int i = 0; i < m; i++) {
      scanf("%d%d%lld",&u,&v,&w);
      e[i].u = u; e[i].v = v; e[i].w = w;
    }
    sort(e, e+m, cmp);
    for(int i = 0; i < m; i++) {
      u = e[i].u; v = e[i].v; w = e[i].w;
      if(unite(u, v)) {
        ret1 += w;
        adde(u, v, w);
      }
    }
    dfs(1, -1);
    LL k = n * (n - 1);
    printf("%lld %.2lf
", ret1, 2.0*ret2/(double)k);
  }
  return 0;
}