[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 = ;
 const int maxm = ;
 int n, m;
 E e[maxm<<];
 Edge edge[maxm<<];
 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 ;
   }
   return ;
 }
 
 void init() {
   memset(head, -, sizeof(head));
   ecnt = ; ret1 = ; ret2 = ;
   for(int i = ; 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 = ;
   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 = ; 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 = ; 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(, -);
     LL k = n * (n - );
     printf("%lld %.2lf\n", ret1, 2.0*ret2/(double)k);
   }
   return ;
 }