HDU_2376
对于任意一棵子树来讲，以根节点为深度最浅的点的路径一共有两类，一类是以根节点为端点的路径，另一类是过根节点但端点分别在两棵子树中的路径。然后将无根树转化为有根树后dfs时计算出上面两类路径的长度即可。
#include<stdio.h> #include<string.h> #include<algorithm> #include<queue> #define MAXD 10010 #define MAXM 20010 typedef long long LL; LL f[MAXD], ANS; int N, first[MAXD], size[MAXD], e, next[MAXM], v[MAXM], w[MAXM]; void add(int x, int y, int z) {v[e] = y, w[e] = z;next[e] = first[x], first[x] = e ++; } void init() {int i, x, y, z;scanf("%d", &N);memset(first, -1, sizeof(first[0]) * N), e = 0;for(i = 1; i < N; i ++){scanf("%d%d%d", &x, &y, &z);add(x, y, z), add(y, x, z);} } struct St {LL f;int size;St(){}St(LL _f, int _size) : f(_f), size(_size){} }; void dfs(int cur, int fa) {int i, s = 0;LL A = 0;size[cur] = 1, f[cur] = 0;std::queue <St> q;for(i = first[cur]; i != -1; i = next[i])if(v[i] != fa){dfs(v[i], cur);size[cur] += size[v[i]], f[cur] += f[v[i]] + w[i] * size[v[i]];q.push(St(f[v[i]] + size[v[i]] * w[i], size[v[i]]));}ANS += f[cur];while(!q.empty()){St st = q.front();q.pop();ANS += st.size * A + s * st.f;s += st.size, A += st.f;} } void solve() {ANS = 0;dfs(0, -1);printf("%.7f\n", (double)ANS * 2 / (N * (N - 1))); } int main() {int t;scanf("%d", &t);while(t --){init();solve();}return 0; }
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