如果k==1， 显然就是直径。
k==2的时候， 把直径的边权变为-1， 然后在求一次直径。 变为-1是因为如果在走一次这条边， 答案会增加1.
学到了新的求直径的方法...
#include <bits/stdc++.h> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, n, a) for(int i = a; i<n; i++) #define fi first #define se second typedef complex <double> cmx; typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; const int maxn = 1e5+5; int head[maxn], num, mx1[maxn], mx2[maxn], n, maxx, p; struct node { int to, nextt, val; }e[maxn*2]; void add(int u, int v, int val) { e[num].to = v, e[num].nextt = head[u], e[num].val = val, head[u] = num++; } void init() { num = 0; mem1(head); } int dfs(int u, int fa) { int maxx1 = 0, maxx2 = 0; for(int i = head[u]; ~i; i = e[i].nextt) { int v = e[i].to; if(v == fa) continue; int tmp = e[i].val + dfs(v, u); if(tmp > maxx1) { maxx2 = maxx1; maxx1 = tmp; mx2[u] = mx1[u]; mx1[u] = i; } else if(tmp > maxx2) { maxx2 = tmp; mx2[u] = i; } } if(maxx < maxx1 + maxx2) { maxx = maxx1+maxx2; p = u; } return maxx1; } int main() { int k, u, v; cin>>n>>k; init(); for(int i = 0; i < n - 1; i ++) { scanf("%d%d", &u, &v); add(u, v, 1); add(v, u, 1); } dfs(1, 0); int ans = 2*(n-1); ans -= (maxx - 1); if(k == 2) { for(int i = mx1[p]; i; i = mx1[e[i].to]) e[i].val = e[i^1].val = -1; for(int i = mx2[p]; i; i = mx1[e[i].to]) e[i].val = e[i^1].val = -1; maxx = 0; dfs(1, 0); ans -= (maxx-1); } cout<<ans<<endl; return 0; }