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P2056 [ZJOI2007] 捉迷藏 题解

P2056 [ZJOI2007] 捉迷藏 题解

题面

题目链接

解法

想到是点分树就比较好做了。

看到树上路径想到$LCA$或点分树。先考虑单次查询的点分治做法,$dp1_u$表示$u$为重心的块中到$fa_u$的最长距离,$dp0_u$表示$u$为重心的块中经过$u$的最长路径,显然$dp0_u$为每个儿子$v$的$dp1_v$最大值与次大值之和。

将此方法拓展到点分树上,我们可以为每个节点维护两个可删堆,分别就表示$dp0_u$和$dp1_u$的值,最后统计到答案中。

空间复杂度$O(n \log n)$,时间复杂度$O(n \log^2 n)$。

AC代码

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/**
* @file: P2056.cpp
* @author: yaoxi-std
* @url:
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
const int MAXN = 1e5 + 10;
const int LOGN = 18;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct heap {
priority_queue<int> q, p;
void push(int x) {
q.push(x);
}
void erase(int x) {
p.push(x);
}
int top() {
while (p.size() && q.size() && p.top() == q.top())
p.pop(), q.pop();
return q.top();
}
void pop() {
while (p.size() && q.size() && p.top() == q.top())
p.pop(), q.pop();
q.pop();
}
int size() {
return q.size() - p.size();
}
int sectop() {
int t = top();
pop();
int r = top();
push(t);
return r;
}
};
int n, q, fa[MAXN], dep[MAXN], fq[MAXN][LOGN];
int scnt, siz[MAXN], mxsiz[MAXN];
bool vis[MAXN], opn[MAXN];
vector<int> g[MAXN];
heap shp, hp[MAXN][2];
char buf[4];
int getroot(int u, int f, int sz) {
int ret = 0;
siz[u] = 1, mxsiz[u] = 0;
for (auto v : g[u]) {
if (v == f || vis[v])
continue;
int tmp = getroot(v, u, sz);
if (mxsiz[tmp] < mxsiz[ret])
ret = tmp;
siz[u] += siz[v];
mxsiz[u] = max(mxsiz[u], siz[v]);
}
mxsiz[u] = max(mxsiz[u], sz - siz[u]);
if (mxsiz[u] < mxsiz[ret])
ret = u;
return ret;
}
void build1(int u, int f) {
fq[u][0] = f, dep[u] = dep[f] + 1;
for (int i = 1; i < LOGN; ++i)
fq[u][i] = fq[fq[u][i - 1]][i - 1];
for (auto v : g[u])
if (v != f)
build1(v, u);
}
void build2(int u, int f, int sz) {
vis[u] = true, fa[u] = f;
for (auto v : g[u]) {
if (v == f || vis[v])
continue;
int ns = (siz[u] > siz[v]) ? siz[v] : sz - siz[u];
int rt = getroot(v, u, ns);
build2(rt, u, ns);
}
}
int lca(int u, int v) {
if (dep[u] < dep[v])
swap(u, v);
int t = dep[u] - dep[v];
for (int i = LOGN - 1; ~i; --i)
if ((t >> i) & 1)
u = fq[u][i];
if (u == v)
return u;
for (int i = LOGN - 1; ~i; --i)
if (fq[u][i] != fq[v][i])
u = fq[u][i], v = fq[v][i];
return fq[u][0];
}
int dist(int u, int v) {
return dep[u] + dep[v] - (dep[lca(u, v)] << 1);
}
void update(int x) {
if (hp[x][0].size() >= 2)
shp.erase(hp[x][0].top() + hp[x][0].sectop());
if (opn[x])
hp[x][0].push(0);
else
hp[x][0].erase(0);
if (hp[x][0].size() >= 2)
shp.push(hp[x][0].top() + hp[x][0].sectop());
for (int u = x; fa[u]; u = fa[u]) {
if (hp[fa[u]][0].size() >= 2)
shp.erase(hp[fa[u]][0].top() + hp[fa[u]][0].sectop());
if (hp[u][1].size())
hp[fa[u]][0].erase(hp[u][1].top());
if (opn[x])
hp[u][1].push(dist(fa[u], x));
else
hp[u][1].erase(dist(fa[u], x));
if (hp[u][1].size())
hp[fa[u]][0].push(hp[u][1].top());
if (hp[fa[u]][0].size() >= 2)
shp.push(hp[fa[u]][0].top() + hp[fa[u]][0].sectop());
}
}
int query() {
return shp.size() ? shp.top() : (scnt ? 0 : -1);
}
signed main() {
read(n);
for (int i = 1; i < n; ++i) {
int u, v;
read(u), read(v);
g[u].emplace_back(v);
g[v].emplace_back(u);
}
mxsiz[0] = INF;
build1(1, 0);
build2(getroot(1, 0, n), 0, n);
for (int i = 1; i <= n; ++i)
opn[i] = true, update(i), ++scnt;
read(q);
while (q--) {
scanf("%s", buf);
if (buf[0] == 'G') {
write(query()), putchar('\n');
} else {
int x;
read(x);
opn[x] ? --scnt : ++scnt;
opn[x] = !opn[x];
update(x);
}
}
return 0;
}