为什么编号改为从 $1$ ~ $n + q$ 就能过

正确代码

#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
const int N = 5e5 + 5;
const int V = 1e6 + 5;
const int inf = 0x3f3f3f3f;
struct Node {
	int x, y, opt, id;
} a[N << 1], tmp[N << 1], b[N << 1];
int n, q, tot, len, ans[N << 1];
struct bit {
	int tr[V];
	inline int lowbit(int x) {
		return x & -x;
	}
	inline void update(int p, int v) {
		while (p <= len) {
			tr[p] = max(tr[p], v);
			p += lowbit(p);
		}	
	}
	inline int query(int x) {
		int res = -inf;
		while (x) {
			res = max(res, tr[x]);
			x -= lowbit(x);
		}
		return res;
	}
	inline void clear(int x) {
		while (x <= len) {
			tr[x] = -inf;
			x += lowbit(x);
		}
	}
	void init() {
		for (int i = 1; i <= len; i++) tr[i] = -inf;
	}
} T;
inline void cdq(int l, int r) {
	if (l == r) return ;
	int mid = (l + r) >> 1;
	cdq(l, mid), cdq(mid + 1, r);
	int p = l, q = mid + 1, pos = l;
	while (p <= mid && q <= r) {
		if (b[p].x <= b[q].x) {
			if (b[p].opt == 1) T.update(b[p].y, b[p].x + b[p].y);
			tmp[pos++] = b[p++];
		} else {
			if (b[q].opt == 2) ans[b[q].id] = min(ans[b[q].id], b[q].x + b[q].y - T.query(b[q].y));
			tmp[pos++] = b[q++];
		}
	}
	while (p <= mid) tmp[pos++] = b[p++];
	while (q <= r) {
		if (b[q].opt == 2) ans[b[q].id] = min(ans[b[q].id], b[q].x + b[q].y - T.query(b[q].y));
		tmp[pos++] = b[q++];
	}
	for (int i = l; i <= mid; i++) if (b[i].opt == 1) T.clear(b[i].y);
	for (int i = l; i <= r; i++) b[i] = tmp[i];
}
inline void solve(int rx, int ry) {
	T.init();
	for (int i = 1; i <= tot; i++) {
		b[i] = a[i];
		if (rx) b[i].x = len - b[i].x;
		if (ry) b[i].y = len - b[i].y;
	}
	cdq(1, tot);
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n >> q;
	for (int i = 1, x, y; i <= n; i++) {
		cin >> x >> y;
		a[++tot] = {++x, ++y, 1, i};// 这里是i
		len = max(len, max(x, y));
	} 
	for (int i = n + 1, opt, x, y; i <= n + q; i++) {  //从n + 1 ~ n + q 
		cin >> opt >> x >> y;
		a[++tot] = {++x, ++y, opt, i};
		len = max(len, max(x, y));
		ans[i] = inf;
	}
	len++;
	solve(0, 0), solve(0, 1), solve(1, 0), solve(1, 1);
	for (int i = 1; i <= tot; i++) if (a[i].opt == 2) cout << ans[i] << endl;
	return 0;
}

原代码

#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
const int N = 5e5 + 5;
const int V = 1e6 + 5;
const int inf = 0x3f3f3f3f;
struct Node {
	int x, y, opt, id;
} a[N << 1], tmp[N << 1], b[N << 1];
int n, q, tot, len, ans[N << 1];
struct bit {
	int tr[V];
	inline int lowbit(int x) {
		return x & -x;
	}
	inline void update(int p, int v) {
		while (p <= len) {
			tr[p] = max(tr[p], v);
			p += lowbit(p);
		}	
	}
	inline int query(int x) {
		int res = -inf;
		while (x) {
			res = max(res, tr[x]);
			x -= lowbit(x);
		}
		return res;
	}
	inline void clear(int x) {
		while (x <= len) {
			tr[x] = -inf;
			x += lowbit(x);
		}
	}
	void init() {
		for (int i = 1; i <= len; i++) tr[i] = -inf;
	}
} T;
inline void cdq(int l, int r) {
	if (l == r) return ;
	int mid = (l + r) >> 1;
	cdq(l, mid), cdq(mid + 1, r);
	int p = l, q = mid + 1, pos = l;
	while (p <= mid && q <= r) {
		if (b[p].x <= b[q].x) {
			if (b[p].opt == 1) T.update(b[p].y, b[p].x + b[p].y);
			tmp[pos++] = b[p++];
		} else {
			if (b[q].opt == 2) ans[b[q].id] = min(ans[b[q].id], b[q].x + b[q].y - T.query(b[q].y));
			tmp[pos++] = b[q++];
		}
	}
	while (p <= mid) tmp[pos++] = b[p++];
	while (q <= r) {
		if (b[q].opt == 2) ans[b[q].id] = min(ans[b[q].id], b[q].x + b[q].y - T.query(b[q].y));
		tmp[pos++] = b[q++];
	}
	for (int i = l; i <= mid; i++) if (b[i].opt == 1) T.clear(b[i].y);
	for (int i = l; i <= r; i++) b[i] = tmp[i];
}
inline void solve(int rx, int ry) {
	T.init();
	for (int i = 1; i <= tot; i++) {
		b[i] = a[i];
		if (rx) b[i].x = len - b[i].x;
		if (ry) b[i].y = len - b[i].y;
	}
	cdq(1, tot);
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n >> q;
	for (int i = 1, x, y; i <= n; i++) {
		cin >> x >> y;
		a[++tot] = {++x, ++y, 1, 0};//这里改为0
		len = max(len, max(x, y));
	} 
	for (int i = 1, opt, x, y; i <= q; i++) { // 从 1 ~ q
		cin >> opt >> x >> y;
		a[++tot] = {++x, ++y, opt, i};
		len = max(len, max(x, y));
		ans[i] = inf;
	}
	len++;
	solve(0, 0), solve(0, 1), solve(1, 0), solve(1, 1);
	for (int i = 1; i <= tot; i++) if (a[i].opt == 2) cout << ans[i] << endl;
	return 0;
}