给出个节点的一棵树，每个节点有开和关两种状态，且每个点有权为，给出个操作，每次改变一个节点的状态(给出负数则关，正数则开)，求每次操作过后，有多少个节点满足，自己处于开状态，且子树内处于关状态的节点数严格大于该节点的.
时限五秒
Analysis
考虑到每次修改，只会影响到其到根节点的节点对答案的贡献，每一次操作可以看成对一个节点的加减，求的是有多少个节点的。这个东西用数据结构及其难维护。再分析一下，每一次操作影响到的点仅是一部分点，很多点的贡献未变。那么是否可以一次计算多次操作的答案。我们可以把操作分块，每个操作放到一起做。具体做法是:考虑修改的节点提出来，建成虚树，那么虚树上每个儿子到父亲的边上的原树的点，对于一次影响它们的操作，他们的变化量是相同的。那么可以把它们拉出来排个序，去重，每条边维护一个指针，指向它们的零分界线，就可以算出答案(注意点的开关状态)，修改点的答案要单独算。复杂度约为，常数比较大，排序的时候用基数排序才能达到准确的复杂度，由于偷懒，用了快排。
# include<cstdio> # include<cstring> # include<algorithm> # include<cmath> # include<vector> using namespace std; const int N = 1e5 + 5; int sta[N],dfn[N],fa[N][25],dep[N],p[N][3]; int t[N],b[N],a[N],e[N],vis[N],col[N],out[N]; int st[N],to[N << 1],nx[N << 1],ans[N],q[N],f[N]; int n,m,tot,len,cnt,h; vector <int> inl[N]; bool cmp(int x,int y) { return dfn[x] < dfn[y]; } bool cmp1(int x,int y) { return t[x] < t[y]; } void add(int u,int v) { to[++tot] = v,nx[tot] = st[u],st[u] = tot; to[++tot] = u,nx[tot] = st[v],st[v] = tot; } void dfs(int u) { dfn[u] = ++tot,dep[u] = dep[fa[u][0]] + 1; for (int i = 1 ; i <= 18 ; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1]; for (int i = st[u] ; i ; i = nx[i]) if (to[i] != fa[u][0]) dfs(to[i]); out[u] = tot; } inline int getlca(int x,int y) { if (dep[x] < dep[y]) swap(x,y); for (int i = 18 ; ~i ; --i) if (dep[fa[x][i]] >= dep[y]) x = fa[x][i]; if (x == y) return x; for (int i = 18 ; ~i ; --i) if (fa[x][i] != fa[y][i]) x = fa[x][i],y = fa[y][i]; return fa[x][0]; } inline void build(int l,int r) { memset(f,0,sizeof(f)),memset(col,0,sizeof(col)); int top = 0; b[h = 1] = 1; for (int i = l ; i <= r ; ++i) b[++h] = q[i]; sort(b + 1,b + h + 1); h = unique(b + 1,b + h + 1) - b - 1; sort(b + 1,b + h + 1,cmp); int h1 = h; for (int i = 1 ; i <= h; ++i) { if (!top) { sta[++top] = b[i]; continue; } int x = b[i],lca = getlca(x,sta[top]); while (dfn[lca] < dfn[sta[top]]) { if (dfn[sta[top - 1]] <= dfn[lca]) { f[sta[top--]] = lca; if (sta[top] != lca) sta[++top] = lca,b[++h1] = lca; break; }else f[sta[top]] = sta[top - 1],--top; } sta[++top] = x; } while (top > 1) f[sta[top]] = sta[top - 1],--top; h = h1; sort(b + 1,b + h + 1,cmp); for (int i = 1 ; i <= h ; ++i) col[b[i]] = 1; } inline void solve(int l,int r) { for (int i = 1 ; i <= n ; ++i) a[i] = i,e[i] = 0; sort(a + 1,a + n + 1,cmp1); for (int i = 1 ; i <= h ; ++i) for (int j = fa[b[i]][0] ; j != f[b[i]] ; j = fa[j][0]) e[j] = b[i]; int ret = 0; for (int i = 1 ; i <= n ; ++i) if (e[a[i]]) inl[e[a[i]]].push_back(a[i]); else ret += (!vis[a[i]] && !col[a[i]] && t[a[i]] < 0); for (int i = l ; i <= r ; ++i) ans[i] = ret; for (int i = 1 ; i <= h ; ++i) { int ed = 0,x = b[i]; for (int j = 0 ; j < inl[x].size() ; ++j) if (ed && t[inl[x][j]] == t[inl[x][j - 1]]) p[ed][1] += (!vis[inl[x][j]]); else p[++ed][0] = inl[x][j],p[ed][1] = (!vis[inl[x][j]]); int p1 = 0,tag = 0,sum = 0; while (p1 < ed && t[p[p1 + 1][0]] < 0) ++p1,sum += p[p1][1]; for (int j = l ; j <= r ; ++j) { vis[q[j]] ^= 1; if (dfn[q[j]] >= dfn[x] && dfn[q[j]] <= out[x]) { t[x] += (vis[q[j]]) ? -1 : 1; if (vis[q[j]]) --tag; else ++tag; } while (p1 && t[p[p1][0]] + tag >= 0) sum -= p[p1][1],--p1; while (p1 < ed && t[p[p1 + 1][0]] + tag < 0) ++p1,sum += p[p1][1]; ans[j] += sum; if (t[x] < 0 && !vis[x]) ++ans[j]; } for (int j = l ; j <= r ; ++j) vis[q[j]] ^= 1; for (int j = 0 ; j < inl[x].size() ; ++j) t[inl[x][j]] += tag; } for (int i = l ; i <= r ; ++i) vis[q[i]] ^= 1; for (int i = 1 ; i <= h ; ++i) inl[b[i]].clear(); } int main() { scanf("%d%d",&n,&m); len = (int)sqrt(m * 15); for (int i = 2 ; i <= n ; ++i) scanf("%d",&fa[i][0]),add(i,fa[i][0]); for (int i = 1 ; i <= n ; ++i) scanf("%d",&t[i]); for (int i = 1 ; i <= m ; ++i) scanf("%d",&q[i]),q[i] = abs(q[i]); tot = 0,dfs(1); cnt = m / len + ((m % len) ? 1 : 0); for (int i = 1 ; i <= cnt ; ++i) { int l = (i - 1) * len + 1,r = min(m,i * len); build(l,r); solve(l,r); } for (int i = 1 ; i <= m ; ++i) printf("%d ",ans[i]); return 0; }