Luogu P5838 P5838 [USACO19DEC]Milk Visits G(树链剖分,二分)
文章目录
题目描述
给定一棵树,$M$ 次询问,树上两点之间的简单路径上是否有某种颜色的节点。
$1 \le N \le 10^5$
$1 \le M \le 10^5$
简要做法
这题是树!先剖了再说
那么问题就变为了:在跳重链的时候,看看当前链上有没有颜色为 $k$ 的点
看起来就很不优美,事实上对于每次操作我们只关注那种颜色的点
所以,把每种颜色的点的 dfn 序开个 vector 存起来
那么就变成了一个经典问题:数组中是否存在一个数 $x$,满足 $dfn[top[u]] \le x \le dfn[u]$
二分即可
最坏时间复杂度 $O(n{log}^{2}n)$
参考代码
#include <stdio.h>
#include <algorithm>
#include <memory.h>
#include <vector>
const int N = 1e5 + 5;
const int M = N << 1;
int n, m;
int a[N];
int head[N], num_edge;
int fa[N], son[N], top[N], dfn[N], idx[N], depth[N], size[N];
std::vector<int> v[N];
struct Node
{
int next, to;
} edge[M];
void add_edge(int u, int v) { edge[++num_edge] = Node{head[u], v}, head[u] = num_edge; }
int read()
{
int x = 0, f = 1;
char ch = getchar();
while ('0' > ch or ch > '9')
f = ch == '-' ? -1 : 1, ch = getchar();
while ('0' <= ch and ch <= '9')
x = x * 10 + ch - 48, ch = getchar();
return x * f;
}
void dfs1(int u, int fa)
{
::fa[u] = fa, size[u] = 1, depth[u] = depth[fa] + 1;
for (int i = head[u], v; i; i = edge[i].next)
if ((v = edge[i].to) != fa)
dfs1(v, u), size[u] += size[v], son[u] = size[v] > size[son[u]] ? v : son[u];
}
void dfs2(int u)
{
dfn[u] = ++dfn[0], idx[dfn[u]] = u, top[u] = u == son[fa[u]] ? top[fa[u]] : u;
if (son[u])
dfs2(son[u]);
for (int i = head[u], v; i; i = edge[i].next)
if ((v = edge[i].to) != fa[u] and v != son[u])
dfs2(v);
}
int main()
{
n = read(), m = read();
for (int i = 1; i <= n; i++)
a[i] = read();
for (int i = 1, u, v; i < n; i++)
u = read(), v = read(), add_edge(u, v), add_edge(v, u);
dfs1(1, 0), dfs2(1);
for (int i = 1; i <= n; i++)
v[a[idx[i]]].push_back(i);
for (int x, y, k; m--;)
{
x = read(), y = read(), k = read();
bool flag = false;
for (; top[x] != top[y]; x = fa[top[x]])
{
if (dfn[top[x]] < dfn[top[y]])
std::swap(x, y);
std::vector<int>::iterator it = std::lower_bound(v[k].begin(), v[k].end(), dfn[top[x]]);
if (it != v[k].end() and *it <= dfn[x])
{
flag = true;
break;
}
}
if (!flag)
{
if (dfn[x] > dfn[y])
std::swap(x, y);
std::vector<int>::iterator it = std::lower_bound(v[k].begin(), v[k].end(), dfn[x]);
if (it != v[k].end() and *it <= dfn[y])
flag = true;
}
printf("%d", flag);
}
return 0;
}