Blog of RuSun

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

P4630 [APIO2018] 铁人两项

P4630 [APIO2018] 铁人两项

考虑对于树如何计算,对于每一个中转点,其所有不同子树内的两个点均可成为答案。对原图建出圆方树,对于每一个方点,代表了这个点双的所有点,计算过一次答案后,圆点再计算答案就会算重。那么将原点的 $size$ 设为 $-1$ 即可。

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);
}
typedef long long LL;
const int N = 1e5 + 10;
LL ans;
int top, stk[N];
int stmp, cnt, dfn[N], low[N];
int n, m, sz, tot, w[N << 1], f[N << 1];
vector <int> g[N], e[N << 1];
void tarjan (int u)
{
++sz;
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])
{
w[++tot] = 0;
int x;
do
{
x = stk[top--];
e[tot].pb(x), e[x].pb(tot);
++w[tot];
} while (x ^ v);
e[tot].pb(u), e[u].pb(tot), ++w[tot];
}
}
else low[u] = min(low[u], dfn[v]);
}
void dfs (int u, int fa)
{
f[u] = u <= n;
for (int v : e[u]) if (v ^ fa)
{
dfs(v, u);
ans += 2ll * w[u] * f[u] * f[v];
f[u] += f[v];
}
ans += 2ll * w[u] * f[u] * (sz - f[u]);
}
int main ()
{
read(n, m);
for (int i = 1; i <= n; ++i) w[i] = -1;
tot = n;
for (int a, b; m; --m)
read(a, b), g[a].pb(b), g[b].pb(a);
for (int i = 1; i <= n; ++i) if (!dfn[i])
{
sz = 0;
tarjan(i), --top;
dfs(i, 0);
}
write(ans);
return 0;
}