http://47.96.116.66/problem.php?cid=1177&pid=8

http://47.96.116.66/problem.php?cid=1180&pid=8

題解：LCA+線段樹

來自：https://blog.csdn.net/kzn2683331518/article/details/88807519

#include <bits/stdc++.h> #define ll long long using namespace std; namespace _{char buf[100000], *p1 = buf, *p2 = buf; bool rEOF = 1;//為0表示文件結(jié)尾inline char nc(){ return p1 == p2 && rEOF && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? (rEOF = 0, EOF) : *p1++; }template<class _T>inline void read(_T &num){char c = nc(), f = 1; num = 0;while (c<'0' || c>'9')c == '-' && (f = -1), c = nc();while (c >= '0'&&c <= '9')num = num * 10 + c - '0', c = nc();num *= f;}inline bool need(char &c){ return c >= 'a'&&c <= 'z' || c >= '0'&&c <= '9' || c >= 'A'&&c <= 'Z'; }//讀入的字符范圍inline void read_str(char *a){while ((*a = nc()) && !need(*a) && rEOF); ++a;while ((*a = nc()) && need(*a) && rEOF)++a; --p1, *a = '\0';}template<class _T>inline void println(_T x){static int buf[30], len; len = 0; do buf[len++] = x % 10, x /= 10; while (x);while (len) putchar(buf[--len] + 48);putchar('\n');} }using namespace _; const ll INF = 0x3f3f3f3f3f3f3f3f; const ll mod = 1e9+7; const int MAXN = 500005;int n, q, k; int a[MAXN];struct edge{int v,next; }e[MAXN<<1]; int head[MAXN], cnt; inline void add(int u,int v){e[cnt] = (edge){v,head[u]}, head[u] = cnt++; }int dep[MAXN], low[MAXN], dfn[MAXN], tot, fa[MAXN][20]; void dfs(int u,int f){dep[u] = dep[f] + 1, dfn[u] = ++tot, fa[u][0] = f;for(int i = 1;i<20;i++)fa[u][i] = fa[fa[u][i-1]][i-1];for(int i = head[u];~i;i = e[i].next){if(e[i].v == f)continue;dfs(e[i].v, u);}low[u] = tot; } inline int lca(int u,int v){if(dep[u]<dep[v])swap(u,v);for(int i = 0, del = dep[u]-dep[v];del;i++, del>>=1)if(del&1)u = fa[u][i];if(u == v)return u;for(int i = 19;i>=0;i--)if(fa[u][i]!=fa[v][i])u = fa[u][i], v = fa[v][i];return fa[u][0]; } #define mid ((l+r)>>1) struct Tree{int lazy, sum; }T[MAXN<<2]; inline void pushdown(int u,int l,int r){if(~T[u].lazy){T[u<<1].lazy = T[u<<1|1].lazy = T[u].lazy;T[u<<1].sum = T[u].lazy*(mid-l+1);T[u<<1|1].sum = T[u].lazy*(r-mid);T[u].lazy = -1;} } void update(int u,int l,int r,int L,int R,int x){if(L<=l&&r<=R){T[u].lazy = x;T[u].sum = x*(r-l+1);return;}pushdown(u,l,r);if(L<=mid)update(u<<1,l,mid,L,R,x);if(R>mid)update(u<<1|1,mid+1,r,L,R,x);T[u].sum = T[u<<1].sum + T[u<<1|1].sum; } int main() {memset(head,-1,sizeof head);read(n), read(q);for(int i = 1,u,v;i<n;i++){read(u), read(v), add(u,v), add(v,u);}dfs(1,1);while(q--){read(k); for(int i = 1;i<=k;i++) read(a[i]);if(k == 1){println(n); continue;}T[1].lazy = 0;int rt = a[1];for(int i = 2;i<=k;i++){if(dfn[rt] <= dfn[a[i]] && dfn[a[i]] <= low[rt]){ //情況1,在rt的子樹中int d = ((dep[a[i]] + dep[rt])>>1) + 1, u = a[i];for(int j = 19;j>=0;j--)if(dep[fa[u][j]] >= d) u = fa[u][j];update(1,1,n,dfn[u],low[u],1); //對u整個子樹打標(biāo)記}else{int lc = lca(rt, a[i]);int dis = ((dep[a[i]] + dep[rt] - (dep[lc]<<1))>>1)+1;if(dep[a[i]] >= dep[rt]){ //情況2int d = dis - dep[rt] + (dep[lc]<<1), u = a[i];for(int j = 19;j>=0;j--)if(dep[fa[u][j]] >= d) u = fa[u][j];update(1,1,n,dfn[u],low[u],1); //對u整個子樹打標(biāo)記}else{ //情況3int d = dep[rt] - dis + 1, u = rt; //d表示u的深度for(int j = 19;j>=0;j--)if(dep[fa[u][j]] >= d) u = fa[u][j];if(1 <= dfn[u]-1)update(1,1,n,1,dfn[u]-1,1); //對u子樹外的點打標(biāo)記if(low[u]+1 <= n)update(1,1,n,low[u]+1,n,1); //對u子樹外的點打標(biāo)記}}}println(n-T[1].sum); //答案為總點數(shù)-被標(biāo)記點數(shù)}return 0; }

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