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给出一棵树，删除一些边，使得任意联通块内的任意点距离不超过$k$
考场上想的贪心是对的：考虑一棵子树，如果该子树内最深的两个节点的距离相加$>k$就删掉最深的那个点，向上update的时候只返回最深的点的深度
然而却苦于写不出代码。。。
/**/#include #include #include #include #include #include #include #include #include //#include //#include #define Pair pair #define MP(x, y) make_pair(x, y)#define fi first#define se second#define int long long #define LL long long #define ull unsigned long long #define rg register #define pt(x) printf("%d ", x);//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)//char buf[(1 << 22)], *p1 = buf, *p2 = buf;//char obuf[1<<24], *O = obuf;//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}//#define OS *O++ = ' ';using namespace std;//using namespace __gnu_pbds;const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;const double eps = 1e-9;inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f;}int N, K, ans = 0;vector v[MAXN];int dfs(int x, int fa) { if(v[x].size() == 1) return 0; vector dis; for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == fa) continue; dis.push_back(dfs(to, x) + 1); } sort(dis.begin(), dis.end()); while(dis.size() >= 2) { int x = dis[dis.size() - 1], y = dis[dis.size() - 2]; if(x + y > K) ans++, dis.pop_back(); else break; } return dis.back();}main() { N = read(); K = read(); for(int i = 1; i <= N - 1; i++) { int x = read(), y = read(); v[x].push_back(y); v[y].push_back(x); } for(int i = 1; i <= N; i++) if(v[i].size() > 1) {dfs(i, 0); break;} printf("%d", ans + 1); return 0;}/*2 2 11 12 1 1*/
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