P6098 [USACO19FEB] Cow Land G

P6098 [USACO19FEB] Cow Land G

可恶，上一个是线段树调试失败，利用树状数组做的，这一次看到单点修改区间查询，果断选着树状数组，结果wa了。
果然我还是太菜了

此题几乎是模板题，只要修改下线段树的向上传递即可，因为是单点修改，所以可以不用懒标记
注意这里求的是异或

点击查看代码

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n;
int a[N];
//树链
vector<int> g[N];
int dep[N], fa[N], sz[N], son[N];
void dfs1(int u,int father) {
	dep[u] = dep[father] + 1;
	fa[u] = father; sz[u] = 1;
	for (int v : g[u]) {
		if (v == father) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v]) son[u] = v;
	}
}
int top[N], cnt, id[N],nw[N];
void dfs2(int u, int t) {
	top[u] = t; id[u] = ++cnt, nw[cnt] = a[u];
	if (!son[u]) return;
	dfs2(son[u], t);
	for (int v : g[u]) {
		if (v == fa[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
//线段树
#define lc k<<1
#define rc k<<1|1
struct tree {
	int l, r;
	int sum;
}t[N*4];
void pushup(int k) {
	t[k].sum = t[lc].sum ^ t[rc].sum;//异或一下
}
void build(int k, int l, int r) {
	t[k] = { l,r };
	if (l == r) {
		t[k].sum = nw[l];
		return;
	}
	int mid = l + r >> 1;
	build(lc, l, mid); build(rc, mid + 1, r);
	pushup(k);
}
void modify(int k, int x,int z) {
	if (t[k].l == x && t[k].r == x) {
		t[k].sum = z;
		return;
	}
	int mid = t[k].l + t[k].r >> 1;
	if (x <= mid) modify(lc, x, z);
	else modify(rc, x, z);
	pushup(k);
}
int query(int k, int l, int r) {
	if (t[k].l >= l && t[k].r <= r) return t[k].sum;
	int mid = t[k].l + t[k].r >> 1;
	int res = 0;
	if (l <= mid) res ^= query(lc, l, r);
	if (r > mid) res ^= query(rc, l, r);
	return res;
}

int ask_path(int u, int v) {
	int res = 0;
	while (top[u] != top[v]) {
		if (dep[top[u]] < dep[top[v]]) swap(v, u);
		res ^= query(1, id[top[u]], id[u]);
		u = fa[top[u]];
	}
	if (dep[u] < dep[v]) swap(u, v);
	res ^= query(1, id[v], id[u]);
	return res;
}
int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int Q;
	cin >> n>>Q;
	for (int i = 1; i <= n; i++) cin >> a[i];
	for (int i = 1; i < n; i++) {
		int x, y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	dfs1(1, 0);
	dfs2(1, 1);
	build(1, 1, n);
	while (Q--) {
		int op, x, y;
		cin >> op >> x >> y;
		if (op == 1) {
			modify(1, id[x], y);
		}
		else {
			cout << ask_path(x, y) << '\n';
		}
	}
	return 0;
}