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101 lines
2.4 KiB
101 lines
2.4 KiB
#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 2e4 + 10, M = 2e5 + 10;
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//链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int depth[N], f[N][31];
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int dlt[N]; //差分数组
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// 倍增2^k,计算k的办法
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// T = log(n) / log(2) + 1;
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const int T = 25;
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//倍增求 a,b的最近公共祖先
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void bfs(int root) {
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queue<int> q;
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q.push(root);
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depth[root] = 1;
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while (q.size()) {
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int u = q.front();
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q.pop();
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (depth[j]) continue;
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depth[j] = depth[u] + 1;
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f[j][0] = u;
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for (int k = 1; k <= T; k++) f[j][k] = f[f[j][k - 1]][k - 1];
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q.push(j);
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}
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}
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}
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int lca(int a, int b) {
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if (depth[a] < depth[b]) swap(a, b);
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for (int k = T; k >= 0; k--)
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if (depth[f[a][k]] >= depth[b])
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a = f[a][k];
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if (a == b) return a;
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for (int k = T; k >= 0; k--)
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if (f[a][k] != f[b][k])
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a = f[a][k], b = f[b][k];
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return f[a][0];
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}
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//前缀和
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void dfs(int u, int fa) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (j == fa) continue;
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dfs(j, u);
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dlt[u] += dlt[j];
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}
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}
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int main() {
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int T, n, m;
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scanf("%d", &T);
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int cc = 0;
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while (T--) {
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idx = 0;
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memset(h, -1, sizeof h);
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memset(depth, 0, sizeof depth);
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memset(dlt, 0, sizeof dlt);
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scanf("%d%d", &n, &m);
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for (int i = 1; i < n; i++) {
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int a, b;
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scanf("%d%d", &a, &b);
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add(a, b), add(b, a);
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}
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bfs(1);
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for (int i = n; i <= m; i++) {
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int a, b;
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scanf("%d%d", &a, &b);
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//树上差分+边权(把边权记在点上)
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dlt[a]++;
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dlt[b]++;
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dlt[lca(a, b)] -= 2;
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}
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//前缀和合并差分
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dfs(1, 0);
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int res = INF;
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/*
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Q:为什么最后统计的时候 n 是从 2 开始?
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A:因为1没边的, 边差分，他这边是下放到下面的那个点上，用点来表示这个边的。
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*/
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for (int i = 2; i <= n; i++) res = min(res, dlt[i] + 1);
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printf("Case #%d: %d\n", ++cc, res);
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}
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}
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