CodeForces 343D Water Tree（树链剖分+dfs时间戳）

题意：有一棵树，每个节点有一个权值0或者1，有三种操作，第一种是将结点u和它的子树中的结点权值全部置为1，另一种是将结点u及其祖先节点权值全部置为0，然后还有查询操作，查询某个结点的权值。

思路：树链剖分。对于第一种操作，可以利用dfs时间戳记录u子树中最后一个节点在线段树中的位置，然后将子树中所有结点置为1。对于第二种操作就是常规的树链剖分。第三种查询操作只需要在线段树中单点查询即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

const int MAXN = 500500;
int setv[4*MAXN], maxv[4*MAXN];
void pushdown(int o) {
int lc = o*2, rc = o*2 + 1;
if(setv[o] >= 0) { //本节点有标记才传递
setv[lc] = setv[rc] = setv[o];
setv[o] = -1;
}
}
void maintain(int o, int L, int R) {
int lc = 2*o, rc = 2*o + 1;
maxv[o] = 0;
if(R > L) {
maxv[o] = max(maxv[o<<1], maxv[(o<<1)|1]);
}
if(setv[o] >= 0) maxv[o] = setv[o];
}
void update(int o, int L, int R, int y1, int y2, int v) {
int lc = o*2, rc = o*2+1;
if(y1<=L && y2>=R) {
setv[o] = v;
} else {
pushdown(o);
int M = (L+R)>>1;
if(y1 <= M) update(lc, L, M, y1, y2, v); else maintain(lc, L, M);
if(y2 > M) update(rc, M+1, R, y1, y2, v); else maintain(rc, M+1, R);
}
maintain(o, L, R);
}
int query(int o, int L, int R, int y1, int y2) {
if(setv[o] >= 0) return setv[o];
else if(y1<=L && y2>=R) return maxv[o];
else {
int M = (L+R) >> 1, ans = -1;
if(y1 <= M) ans = max(ans, query(o*2, L, M, y1, y2));
if(y2 > M) ans = max(ans, query(o*2+1, M+1, R, y1, y2));
return ans;
}
}
void build(int o, int L, int R) {
setv[o] = 0;
maxv[o] = 0;
if(L == R) return;
int M = (L+R) >> 1;
build(2*o, L, M);
build(2*o+1, M+1, R);
}
int n, tot, dfs_clock, arc[MAXN];
vector<int> G[MAXN];
int siz[MAXN], son[MAXN], dep[MAXN], top[MAXN], fa[MAXN], pos[MAXN];
int tim[MAXN][2];
void init() {
for(int i = 1; i <= n; i++) G[i].clear();
memset(son, 0, sizeof(son));
tot = dfs_clock = 0;
}
void dfs(int cur, int f) {
siz[cur] = 1;
int tmp = 0;
for(int i = 0; i < G[cur].size(); i++) {
int u = G[cur][i];
if(u == f) continue;
dep[u] = dep[cur] + 1;
fa[u] = cur;
dfs(u, cur);
siz[cur] += siz[u];
if(siz[u] > tmp) son[cur] = u, tmp = siz[u];
}
}
void dfs2(int cur, int tp) {
top[cur] = tp;
pos[cur] = ++tot;
tim[cur][0] = ++dfs_clock;
arc[dfs_clock] = cur;
if(son[cur]) dfs2(son[cur], tp);
for(int i = 0; i < G[cur].size(); i++) {
int u = G[cur][i];
if(u==son[cur] || u==fa[cur]) continue;
dfs2(u, u);
}
tim[cur][1] = dfs_clock;
}
void modify(int cur, int v) {
int tp = top[cur];
while(tp != 1) {
update(1, 1, n, pos[tp], pos[cur], v);
cur = fa[tp]; tp = top[cur];
}
update(1, 1, n, pos[tp], pos[cur], v);
}
int main() {
//freopen("input.txt", "r", stdin);
cin >> n;
for(int i = 1, u, v; i < n; i++) {
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1, -1);
dfs2(1, 1);
build(1, 1, n);
int q; cin >> q;

//printf("%d\n", query(1, 1, n, 1, 1));
for(int i = 1, u, v; i <= q; i++) {
scanf("%d%d", &u, &v);
if(u==1) update(1, 1, n, pos[v], pos[arc[tim[v][1]]], 1);
else if(u==2) modify(v, 0);
else printf("%d\n", query(1, 1, n, pos[v], pos[v]));
}
//update(1, 1, n, 1, n, 1);
//for(int i = 1; i <= n; i++) printf("%d\n", pos[i]);
return 0;
}