【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;

}

巴特西

【CF613D】Kingdom and its Cities

【CF613D】Kingdom and its Cities

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题解

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