[10.5模拟赛]T1
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T1
Description
在\(2019\)年,某小朋友刚刚学习了树,非常开心。现在他想解决这样一个问题:给定一颗有根树(根为\(1\)),有以下两种操作:
标记操作:对某个结点打上标记(在最开始,只有结点\(1\)有标记,其他结点均无标记,而且对于某个结点,可以打多次标记。)
询问操作:询问某个结点最近的一个打了标记的祖先(这个结点本身也算自己的祖先)
你能帮帮他吗?
Input
输入第一行两个正整数\(N\)和\(Q\)分别表示节点个数和操作次数
接下来\(N-1\)行,每行两个正整数\(u\),\(v(1≤u\),\(v≤n)\)表示\(u\)到\(v\)有一条有向边
接下来\(Q\)行,形如“\(oper num\)”\(oper\)为“\(C\)”时表示这是一个标记操作,\(oper\)为“\(Q\)”时表示这是一个询问操作对于每次询问操作。
Output
对于每个询问,输出一个正整数,表示结果。
Sample Input
5 5
1 2
1 3
2 4
2 5
Q 2
C 2
Q 2
Q 5
Q 3
Sample Output
1
2
2
1
Data Constraint
$30%的数据,\(1?N,Q?1000\)
\(70\)%的数据,\(1?N,Q?10000\)
\(100\)%的数据,\(1?N,Q?100000\)
Solution
Code
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 800010
struct rec
int nxt, ver;
t[MAXN];
struct Rec
int val;
tree[MAXN];
int n, q, cnt, u, v, x, s;
int head[MAXN], Top[MAXN], Fa[MAXN], h[MAXN], H[MAXN];
char ch;
bool vis[MAXN];
inline int read()
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
inline void add(int u, int v)
t[++cnt].nxt = head[u], t[cnt].ver = v, head[u] = cnt;
void DFS(int u, int fa)
vis[u] = true, Top[u] = fa, h[u] = ++s, H[s] = u;
for (register int i = head[u]; i; i = t[i].nxt)
int v = t[i].ver;
if (!vis[v])
Fa[v] = u, DFS(v, fa);
break;
for (register int i = head[u]; i; i = t[i].nxt)
int v = t[i].ver;
if (!vis[v])
Fa[v] = u, DFS(v, v);
inline void update(int now)
tree[now].val = max(tree[now << 1].val, tree[now << 1 | 1].val);
void Modify(int now, int l, int r, int k)
#define mid ((l + r) >> 1)
if (l == r)
tree[now].val = max(tree[now].val, k);
return;
if (k <= mid) Modify(now << 1, l, mid, k);
else Modify(now << 1 | 1, mid + 1, r, k);
update(now);
#undef mid
int Query(int now, int l, int r, int x, int y)
int ans = 0;
#define mid ((l + r) >> 1)
if (x <= l && r <= y)
return max(ans, tree[now].val);
if (x <= mid) ans = max(ans, Query(now << 1, l, mid, x, y));
if (mid < y) ans = max(ans, Query(now << 1 | 1, mid + 1, r, x, y));
return ans;
#undef mid
int fuck(int now)
while (now)
s = Query(1, 1, n, h[Top[now]], h[now]);
if (s) return H[s];
now = Fa[Top[now]];
int main()
freopen("pa.in", "r", stdin);
freopen("pa.out", "w", stdout);
n = read(), q = read();
for (register int i = 1; i <= n - 1; i++)
u = read(), v = read(), add(u, v), add(v, u);
DFS(1, 1);
Modify(1, 1, n, 1);
while (q--)
ch = getchar();
if (ch == 'Q')
x = read();
printf("%d\n", fuck(x));
if (ch == 'C')
x = read();
Modify(1, 1, n, h[x]);
return 0;
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