Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4244 [SHOI2008]仙人掌图 II

P4244 [SHOI2008]仙人掌图 II

将边分为两类,一类是桥,按照树的方式DP。一类是环。环上任意两个点都可以构成一条路径,其中距离为环上的最短距离。不同的子仙人掌可以构成一个路径。前者类似一个滑动窗口,可以用单调队列优化;后者维护当前最大的深度。

查看代码
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#include <cstdio>
#include <vector>
#define pb push_back
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
const int N = 1e5 + 10;
int hd, tl, q[N];
int n, m, ans, p[N], f[N];
vector <int> g[N];
int stmp, dfn[N], low[N];
void solve (vector <int> &h)
{
int l = h.size(), r = l >> 1;
h.resize(l * 2);
for (int i = 0; i < l; ++i) h[i + l] = h[i];
auto add = [&](int i)
{
while (hd <= tl && h[q[tl]] + q[tl] <= h[i] + i) --tl;
q[++tl] = i;
};
hd = 1, tl = 0;
for (int i = 0; i < r; ++i) add(i);
for (int i = 0; i < l * 2; ++i)
{
if (hd <= tl && q[hd] <= i) ++hd;
if (i + r < l * 2) add(i + r);
if (hd <= tl) ans = max(ans, h[i] + h[q[hd]] + q[hd] - i);
}
}
void tarjan (int u)
{
dfn[u] = low[u] = ++stmp;
for (int v : g[u]) if (v ^ p[u])
if (!dfn[v])
{
p[v] = u;
tarjan(v);
low[u] = min(low[u], low[v]);
if (low[v] > dfn[u])
{
ans = max(ans, f[u] + f[v] + 1);
f[u] = max(f[u], f[v] + 1);
}
}
else low[u] = min(low[u], dfn[v]);
for (int v : g[u]) if (dfn[v] > dfn[u] && p[v] ^ u)
{
vector <int> h; h.pb(0);
for (int t = v; t ^ u; t = p[t]) h.pb(f[t]);
for (int i = 1; i < h.size(); ++i)
ans = max(ans, f[u] + h[i] + min(i, (int)h.size() - i));
for (int i = 1; i < h.size(); ++i)
f[u] = max(f[u], h[i] + min(i, (int)h.size() - i));
solve(h);
}
}
int main ()
{
read(n, m);
for (int k; m; --m)
{
read(k);
for (int a, b = 0; k; --k)
{
read(a);
if (b) g[a].pb(b), g[b].pb(a);
b = a;
}
}
tarjan(1);
write(ans);
return 0;
}