题意
给定\(n\)个软件包，每个软件包都有一个依赖软件包，安装一个软件包必须安装他的依赖软件包，卸载一个软件包必须先卸载所有依赖于它的软件包。给定\(m\)此操作，每次一个操作\(install/unistall\)表示安装或者卸载。
题解
可以通过简单画图看出，在这个树形结构的依赖层次图上，安装一个包相当于安装其到根节点路径上的所有包，删除一个包相当于删除其与其子树的包。用一个重链剖分+线段树处理一下就行了。
#include#include using std::swap;typedef long long ll;const int N = 1e5 + 10;int n, Q, x, ans;int fa[N], dep[N], siz[N], son[N];int top[N], dfn[N], time;int cnt, from[N], to[N], nxt[N];int bui[N << 2], set[N << 2];inline void addEdge(int u, int v) { to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;}void dfs1(int u) { siz[u] = 1, dep[u] = dep[fa[u]] + 1; for (int i = from[u]; i; i = nxt[i]) { int v = to[i]; dfs1(v), siz[u] += siz[v]; if(siz[v] > siz[son[u]]) son[u] = v; }}void dfs2(int u, int t) { dfn[u] = ++time, top[u] = t; if(!son[u]) return ; dfs2(son[u], t); for(int i = from[u]; i; i = nxt[i]) { int v = to[i]; if(v == son[u]) continue; dfs2(v, v); }}void modify(int sl, int sr, int k, int o = 1, int l = 1, int r = n) { int len = r - l + 1; if(l >= sl && r <= sr) { if(k == 1) ans += len - bui[o], bui[o] = len; else ans += bui[o], bui[o] = 0; set[o] = k; return ; } int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1; if(set[o]) { if(set[o] == 1) bui[lc] = (len - (len >> 1)), bui[rc] = (len >> 1); else bui[lc] = bui[rc] = 0; set[lc] = set[rc] = set[o], set[o] = 0; } if(sl <= mid) modify(sl, sr, k, lc, l, mid); if(sr > mid) modify(sl, sr, k, rc, mid + 1, r); bui[o] = bui[lc] + bui[rc];}inline void ins(int x) { int fx = top[x]; while (fx != 1) modify(dfn[fx], dfn[x], 1), x = fa[fx], fx = top[x]; modify(1, dfn[x], 1);}int main () { scanf("%d", &n); for (int i = 2; i <= n; ++i) { scanf("%d", fa + i), ++fa[i]; addEdge(fa[i], i); } dfs1(1), dfs2(1, 1); scanf("%d", &Q); char opt[12]; while(Q--) { scanf("\n%s%d", opt, &x), ans = 0, ++x; if(opt[0] == 'i') ins(x); else modify(dfn[x], dfn[x] + siz[x] - 1, -1); printf("%d\n", ans); } return 0;}