【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_{vin 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
", f[1]); 
	  NXT : 
		for (int i = 1; i <= K; i++) key[a[i]] = 0; 
	} 
	return 0; 
}