ZOJ 3863Paths on the Tree

Description

Edward has a tree with n vertices conveniently labeled with 1,2,…,n.

Edward finds a pair of paths on the tree which share no more than k common vertices. Now Edward is interested in the number of such ordered pairs of paths.

Note that path from vertex a to b is the same as the path from vertex b to a. An ordered pair means (A, B) is different from (B, A) unlessA is equal to B.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers n, k (1 ≤ n, k ≤ 88888). Each of the following n - 1 lines contains two integers a_i, b_i, denoting an edge between vertices a_i and b_i (1 ≤ a_i, b_i ≤ n).

The sum of values n for all the test cases does not exceed 888888.

Output

For each case, output a single integer denoting the number of ordered pairs of paths sharing no more than k vertices.

Sample Input

14 21 22 33 4

Sample Output

path Apaths share 2 vertices with Atotal1-2-3-41-2, 2-3, 3-431-2-31-2, 2-3, 2-3-432-3-41-2-3, 2-3, 3-431-21-2, 1-2-3, 1-2-3-432-31-2-3, 1-2-3-4, 2-3, 2-3-443-41-2-3-4, 2-3-4, 3-43

93

Hint

The number of path pairs that shares no common vertex is 30.

The number of path pairs that shares 1 common vertex is 44.

The number of path pairs that shares 2 common vertices is 19.

这种题对于脑细胞的损耗实在有点大，wa了10发才过，实在是有点疲惫。

树分治，然后每个点统计边的种类，细节处理有点复杂。

#include<cstdio>#include<vector>#include<algorithm>using namespace std;typedef unsigned long long LL;const int low(int x) { return x&-x; }const int maxn = 3e5 + 10;const int INF = 0x7FFFFFFF;int T, n, m, x, y;struct Tree{int ft[maxn], nt[maxn], u[maxn], sz;int vis[maxn], mx[maxn], ct[maxn];LL d[maxn], D[maxn];void clear(int n){mx[sz = 0] = INF;for (int i = 1; i <= n; i++) vis[i] = 0, ft[i] = -1;}void AddEdge(int x, int y){u[sz] = y;nt[sz] = ft[x];ft[x] = sz++;u[sz] = x;nt[sz] = ft[y]; ft[y] = sz++;}int dfs(int x, int fa, int sum){int y = mx[x] = (ct[x] = 1) - 1;for (int i = ft[x]; i != -1; i = nt[i]){if (vis[u[i]] || u[i] == fa) continue;int z = dfs(u[i], x, sum);ct[x] += ct[u[i]];mx[x] = max(mx[x], ct[u[i]]);y = mx[y] < mx[z] ? y : z;}mx[x] = max(mx[x], sum - ct[x]);return mx[x] < mx[y] ? x : y;}int getdep(int x, int fa, int dep){int ans = dep;for (int i = ft[x]; i != -1; i = nt[i]){if (u[i] == fa || vis[u[i]]) continue;ans = max(ans, getdep(u[i], x, dep + 1));}return ans;}LL get(int x, int fa, int dep){LL cnt = 1, ans = 0;for (int i = ft[x]; i != -1; i = nt[i]){if (u[i] == fa) continue;LL y = vis[u[i]] ? mx[u[i]] : get(u[i], x, dep + 1);cnt += y;ans += y*y;}D[dep] += cnt*cnt - ans;if (dep == 1) return ans;return cnt;}LL find(int x){LL ans = 0, sum = 0, tot = 0;int len = getdep(x, -1, 1);if (len + len <= m + 1) return 0;sum = get(x, -1, 1);for (int i = 1; i <= len; i++) d[i] = 0;for (int i = ft[x]; i != -1; i = nt[i]){if (vis[u[i]]) continue;int y = getdep(u[i], x, 2);for (int j = 2; j <= y; j++) D[j] = 0;LL z = get(u[i], x, 2);for (int j = 2; j <= y; j++){LL s = 0;for (int k = min(m + 1 - j, len); k > 0; k -= low(k)) s += d[k];ans += D[j] * (tot - s);}for (int j = 2; j <= y; j++){                                                                                                      for (int k = j; k <= len; k += low(k)) d[k] += D[j];if (j > m) ans += (((LL)n - z)*((LL)n - z) + z*z - sum)*D[j];tot += D[j];}}return ans;}int dfs(int x,int fa){int cnt=1;for (int i=ft[x];i!=-1;i=nt[i]){if (u[i]==fa) continue;cnt+=vis[u[i]]?mx[u[i]]:dfs(u[i],x);}return cnt;}LL work(int x, int sum){int y = dfs(x, -1, sum);LL ans = find(y);vis[y] = 1;for (int i = ft[y]; i != -1; i = nt[i]){if (vis[u[i]]) continue;mx[y] = n-dfs(u[i],y);ans += work(u[i], ct[u[i]] > ct[y] ? sum - ct[y] : ct[u[i]]);}return ans;}}solve;int main(){scanf("%d", &T);while (T--){scanf("%d%d", &n, &m);LL ans = (LL)n * (n + 1) >> 1;solve.clear(n);for (int i = 1; i < n; i++){scanf("%d%d", &x, &y);solve.AddEdge(x, y);}printf("%llu\n", ans*ans - solve.work(1, n));}return 0;}