Blog of RuSun

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

P3320 [SDOI2015]寻宝游戏

P3320 [SDOI2015]寻宝游戏

要求若干个点的连通块的边权和的两倍,类似虚树,有一个结论:将所有点按照 dfn 序排序,答案为 $dis(rnk _ 1, rnk _ 2) + dis(rnk _ 2 + rnk _ 3) + \ldots + dis(rnk _ n + dis _ 1)$ 。

查看代码
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#include <cstdio>
#include <set>
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, M = 2e5 + 10;
LL ans, s[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
int n, m, d[N], p[N], sz[N];
int stmp, dfn[N], rnk[N], son[N], top[N];
void dfs1 (int x = 1)
{
sz[x] = 1;
for (int i = hd[x]; ~i; i = nxt[i]) if (edg[i] ^ p[x])
{
p[edg[i]] = x;
d[edg[i]] = d[x] + 1;
s[edg[i]] = s[x] + wt[i];
dfs1(edg[i]);
sz[x] += sz[edg[i]];
if (sz[edg[i]] > sz[son[x]]) son[x] = edg[i];
}
}
void dfs2 (int x = 1, int t = 1)
{
top[x] = t;
rnk[dfn[x] = ++stmp] = x;
if (!son[x]) return;
dfs2(son[x], t);
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] ^ p[x] && edg[i] ^ son[x]) dfs2(edg[i], edg[i]);
}
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 add (int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
LL dis (int a, int b)
{
if (a == 0 || b == 0 || a == n + 1 || b == n + 1) return 0;
a = rnk[a], b = rnk[b];
int t = lca(a, b);
return s[a] + s[b] - 2 * s[t];
}
int main ()
{
read(n, m);
for (int i = 1; i <= n; ++i) hd[i] = -1;
for (int i = 1, a, b, c; i < n; ++i)
read(a, b, c), add(a, b, c), add(b, a, c);
dfs1(), dfs2();
set <int> c;
c.insert(0), c.insert(n + 1);
for (int t; m; --m)
{
read(t); t = dfn[t];
int a = *--c.lower_bound(t), b = *c.upper_bound(t);
if (c.count(t))
{
ans -= dis(a, t) + dis(b, t) - dis(a, b);
c.erase(t);
}
else
{
ans += dis(a, t) + dis(b, t) - dis(a, b);
c.insert(t);
}
write(ans + dis(*++c.begin(), *----c.end())), puts("");
}
return 0;
}