【CF613D】Kingdom and its Cities

【CF613D】Kingdom and its Cities

题面

洛谷

题解

看到关键点当然是建虚树啦。

设\(f[x]\)表示以\(x\)为根的子树的答案，\(g[x]\)表示以\(x\)为根的子树内是否有和\(x\)联通的点，\(c=\sum_{v\in son_x} g[v]\)。

分类讨论一下：

• 如果一个点在原树下它和它的父亲均为关键点，那么显然不行。
• 这个点是关键点，\(f[x]+=c\)，\(g[x]=1\)
• 这个点不是关键点，若\(c>1\)则断掉这个点，\(++f[x],g[x]=0\)，否则\(g[x]=1\)

这题就做\(v.a.n\)啦。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 1e5 + 5;
struct Graph { int to, next; } e[MAX_N << 1];
int fir1[MAX_N], fir2[MAX_N], e_cnt;
void clearGraph() {
	memset(fir1, -1, sizeof(fir1));
	memset(fir2, -1, sizeof(fir2));
}
void Add_Edge(int *fir, int u, int v) {
	e[e_cnt] = (Graph){v, fir[u]};
	fir[u] = e_cnt++;
}
namespace Tree {
    int fa[MAX_N], dep[MAX_N], size[MAX_N], top[MAX_N], son[MAX_N], dfn[MAX_N], tim;
    void dfs1(int x) {
        dfn[x] = ++tim;
        size[x] = 1, dep[x] = dep[fa[x]] + 1;
        for (int i = fir1[x]; ~i; i = e[i].next) {
            int v = e[i].to; if (v == fa[x]) continue;
            fa[v] = x; dfs1(v); size[x] += size[v];
            if (size[v] > size[son[x]]) son[x] = v;
        }
    }
    void dfs2(int x, int tp) {
        top[x] = tp;
        if (son[x]) dfs2(son[x], tp);
        for (int i = fir1[x]; ~i; i = e[i].next) {
            int v = e[i].to; if (v == fa[x] || v == son[x]) continue;
            dfs2(v, v);
        }
    }
    int LCA(int x, int y) {
        while (top[x] != top[y]) {
            if (dep[top[x]] < dep[top[y]]) swap(x, y);
            x = fa[top[x]];
        }
        return dep[x] < dep[y] ? x : y;
    }
}
using Tree::dfn; using Tree::LCA;
int N, M, K, a[MAX_N], f[MAX_N];
bool g[MAX_N], key[MAX_N];
bool cmp(const int &i, const int &j) { return dfn[i] < dfn[j]; }
void build() {
	static int stk[MAX_N], top;
	e_cnt = 0; top = 0;
	sort(&a[1], &a[K + 1], cmp);
	for (int i = 1; i <= K; i++) if (a[i] != a[i - 1]) stk[++top] = a[i];
	K = top;
	for (int i = 1; i <= K; i++) a[i] = stk[i];
	stk[top = 1] = 1, fir2[1] = -1;
	for (int i = 1; i <= K; i++) {
		if (a[i] == 1) continue;
	    int lca = LCA(a[i], stk[top]);
		if (lca != stk[top]) {
			while (dfn[lca] < dfn[stk[top - 1]]) {
				int u = stk[top], v = stk[top - 1];
				Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);
				--top;
			}
			if (dfn[lca] > dfn[stk[top - 1]]) {
				fir2[lca] = -1; int u = lca, v = stk[top];
				Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);
				stk[top] = lca;
			} else {
				int u = lca, v = stk[top--];
				Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);
			}
		}
		fir2[a[i]] = -1, stk[++top] = a[i];
	}
	for (int i = 1; i < top; i++) {
		int u = stk[i], v = stk[i + 1];
		Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);
	}
}
void Dp(int x, int fa) {
	f[x] = g[x] = 0;
	int c = 0;
	for (int i = fir2[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (v == fa) continue;
		Dp(v, x);
		f[x] += f[v], c += g[v];
	}
	if (key[x]) g[x] = 1, f[x] += c;
	else if (c > 1) g[x] = 0, ++f[x];
	else g[x] = c;
}
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
	clearGraph();
	N = gi();
	for (int i = 1; i < N; i++) {
		int u = gi(), v = gi();
		Add_Edge(fir1, u, v), Add_Edge(fir1, v, u);
	}
	Tree::dfs1(1), Tree::dfs2(1, 1);
	M = gi();
	while (M--) {
		K = gi(); for (int i = 1; i <= K; i++) key[a[i] = gi()] = 1;
		bool flg = 1;
		for (int i = 1; i <= K && flg; i++) if (key[Tree::fa[a[i]]]) flg = 0;
		if (!flg) { puts("-1"); goto NXT; }
		build();
		Dp(1, 0);
		printf("%d\n", f[1]);
	  NXT :
		for (int i = 1; i <= K; i++) key[a[i]] = 0;
	}
	return 0;
}