Blog of RuSun

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

P4606 [SDOI2018]战略游戏

P4606 [SDOI2018]战略游戏

求出圆方树,圆方树上的圆点即为割点。现在要求一个连接点集的连通块的路径上的割点的和。类似虚树,按照 dfn 排序,相邻两个求路径。注意到是点权,答案即为边权加上最上面的 lca

查看代码
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#include <cstdio>
#include <vector>
#include <algorithm>
#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 = 2e5 + 10;
int top, stk[N];
vector <int> g[N], e[N];
int stmp, dfn[N], low[N];
int n, m, q, tot, s[N], p[N], d[N], w[N];
void tarjan (int u)
{
stk[++top] = u;
low[u] = dfn[u] = ++stmp;
for (int v : g[u])
if (!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
if (low[v] == dfn[u])
{
int x;
++tot;
do
{
x = stk[top--];
e[tot].pb(x), e[x].pb(tot);
} while (x ^ v);
e[tot].pb(u), e[u].pb(tot);
}
}
else low[u] = min(low[u], dfn[v]);
}
namespace LCA
{
int stmp, dfn[N], sz[N], son[N], top[N];
void dfs1 (int u = 1)
{
sz[u] = 1; son[u] = 0;
for (int v : e[u]) if (v ^ p[u])
{
p[v] = u;
d[v] = d[u] + 1;
s[v] = s[u] + w[v];
dfs1(v);
sz[u] += sz[v];
if (sz[v] > sz[son[u]]) son[u] = v;
}
}
void dfs2 (int u = 1, int t = 1)
{
top[u] = t;
dfn[u] = ++stmp;
if (!son[u]) return;
dfs2(son[u], t);
for (int v : e[u])
if (v ^ p[u] && v ^ son[u]) dfs2(v, v);
}
int lca (int a, int b)
{
while (top[a] ^ top[b])
{
if (d[top[a]] < d[top[b]]) swap(a, b);
a = p[top[a]];
}
if (d[a] > d[b]) swap(a, b);
return a;
}
void init ()
{
stmp = 0;
s[1] = w[1]; dfs1(), dfs2();
}
}
int calc (int a, int b)
{
int t = LCA::lca(a, b);
return s[a] + s[b] - s[t] * 2;
}
int main ()
{
int T; read(T);
while (T--)
{
read(n, m);
for (int a, b; m; --m)
read(a, b), g[a].pb(b), g[b].pb(a);
tot = n;
tarjan(1), --top;
for (int i = 1; i <= n; ++i) w[i] = 1;
for (int i = n + 1; i <= tot; ++i) w[i] = 0;
LCA::init();
read(q);
for (int k; q; --q)
{
read(k);
vector <int> h; h.resize(k);
for (int i = 0; i < k; ++i) read(h[i]);
sort(h.begin(), h.end(), [&](int a, int b) { return LCA::dfn[a] < LCA::dfn[b]; });
int res = calc(h.front(), h.back());
for (int i = 1; i < k; ++i) res += calc(h[i], h[i - 1]);
res >>= 1;
res += w[LCA::lca(h.front(), h.back())];
write(res - k), puts("");
}
for (int i = 1; i <= n; ++i) g[i].clear();
for (int i = 1; i <= tot; ++i) e[i].clear();
stmp = 0;
for (int i = 1; i <= n; ++i) dfn[i] = 0;
}
return 0;
}