Blog of RuSun

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

CF1706E Qpwoeirut and Vertices

LuoGu: CF1706E Qpwoeirut and Vertices

CF: E. Qpwoeirut and Vertices

对于区间连通,考虑对于每个点 $i$ ,$i$ 和 $i + 1$ 点连通的需要多少边,这样区间答案为区间最大值,可以用 ST 表做到 $O(n \log n) / O(1)$ 。现在即有 $n - 1$ 次询问两点间需要多少边连通。考虑求出最小生成树,那么记为树上两点最大值,可以用树上倍增做到 $O(n \log n) / O(\log n)$ 。使用Kruskal重构树可以简化这个过程,求出重构树后,问题转化为求两点 LCA ,可以选择的方法很多。

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 = 4e5 + 10, M = 20;
vector <int> g[N];
int n, tot, m, q, lg[N], p[N], d[N], w[N], v[N][M];
int fa (int x) { return p[x] == x ? x : p[x] = fa(p[x]); }
namespace LCA
{
int stmp, id[N], st[N][M];
int dmin (int a, int b) { return d[a] < d[b] ? a : b; }
void dfs (int x)
{
st[id[x] = ++stmp][0] = x;
for (int i : g[x])
{
d[i] = d[x] + 1;
dfs(i), st[++stmp][0] = x;
}
}
void init ()
{
stmp = 0;
int rt = fa(1); d[rt] = 0; dfs(rt);
for (int k = 0; k < lg[stmp]; ++k)
for (int i = stmp + 1 - (1 << k + 1); i; --i)
st[i][k + 1] = dmin(st[i][k], st[i + (1 << k)][k]);
}
int lca (int a, int b)
{
a = id[a], b = id[b];
if (a > b) swap(a, b);
int k = lg[b - a + 1];
return dmin(st[a][k], st[b - (1 << k) + 1][k]);
}
}
int main ()
{
int T; read(T);
for (int i = 2; i < N; ++i) lg[i] = lg[i >> 1] + 1;
while (T--)
{
read(n, m, q);
tot = n;
for (int i = 1; i <= n; ++i)
p[i] = i, g[i].clear();
for (int i = 1, a, b; i <= m; ++i)
{
read(a, b);
a = fa(a), b = fa(b);
if (a == b) continue;
p[a] = p[b] = ++tot;
p[tot] = tot, g[tot].clear();
w[tot] = i;
g[tot].pb(a), g[tot].pb(b);
}
LCA::init();
for (int i = 1; i < n; ++i)
v[i][0] = w[LCA::lca(i, i + 1)];
for (int k = 0; k < lg[n - 1]; ++k)
for (int i = n - (1 << k + 1); i; --i)
v[i][k + 1] = max(v[i][k], v[i + (1 << k)][k]);
for (int l, r; q; --q)
{
read(l, r);
if (l == r) write(0);
else { int k = lg[r - l]; write(max(v[l][k], v[r - (1 << k)][k])); }
putchar(' ');
}
puts("");
}
return 0;
}