给定两个长度为$n$的数组$a,b$，满足$-10^3\le a_i,b_i\le 10^3$，每个数字$x$表示向前走$x$步，如果是负数则后退嘛，设小$A$执行$a$数组，小$B$执行$b$数组。

有三种操作：

• 把$a$数组中下标为$x$的数修改为$y$。
• 把$b$数组中下标为$x$的数修改为$y$。
• 如果小$A$起始位置在$x$，小$B$起始位置在$y$，问如果他们各自按照数组$a,b$执行，在第几步能够行动轨迹重合。

如何判定行动轨迹重合：

假设小$A$走过的区间为$[l1,r1]$，小$B$走过的区间为$[l2,r2]$，如果重合则有$max(l1,l2)\le min(r1,r2)$，

接下来如何确定小$A$和小$B$走过的区间，设$a$数组的前缀和数组为$sum\_a$，$b$数组的前缀和数组为$sum\_b$，

数组$sum\_a$的前缀最小，前缀最大值数组为$min\_a,max\_a$，数组$sum\_b$的前缀最小，前缀最大值数组为$min\_b,max\_b$，

由小$A$，小$B$的初始位置$x,y$，可以确定$l1=min(x,x+min\_a,r1=max(x,x+max\_a),l2=min(y,y+min\_b,r2=max(y,y+max\_b)$。

所以有一个较为暴力的算法 二分 + 线段树 去$check$答案，当时这样操作复杂度是$O(n{\log }^{2}n)$的，对于$n=10^6$的数据显然是不可行的。

考虑线段树上二分：

假设当前答案在区间为$[l,r]$，如果答案在$[mid+1,r]$上，一定有$[1,mid]$这段区间的前缀最大，前缀最小得到的两个区间是不相交的，

另开四个变量来记录区间$[1,l-1]$中$A$的前缀最小， 最大，$B$的前缀最小，最大，

用这四个变量跟区间$[l,mid]$的最大最小来判断，答案是否落在$[l,mid]$之间，直到递归到叶节点，进行判断是否符合条件即可。

#include <bits/stdc++.h>
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef pair<int, int> pii;const int N = 1e6 + 10;int a[N], b[N], sum[N], n, m;struct Seg_tree {int maxn[N << 2], minn[N << 2], lazy[N << 2];void push_up(int rt) {minn[rt] = min(minn[ls], minn[rs]);maxn[rt] = max(maxn[ls], maxn[rs]);}void push_down(int rt) {if (!lazy[rt]) {return ;}lazy[ls] += lazy[rt], lazy[rs] += lazy[rt];minn[ls] += lazy[rt], minn[rs] += lazy[rt];maxn[ls] += lazy[rt], maxn[rs] += lazy[rt];lazy[rt] = 0;}void build(int rt, int l, int r) {if (l == r) {maxn[rt] = minn[rt] = sum[l];return ;}build(lson);build(rson);push_up(rt);}void update(int rt, int l, int r, int L, int R, int v) {if (l >= L && r <= R) {lazy[rt] += v;minn[rt] += v, maxn[rt] += v;return ;}push_down(rt);if (L <= mid) {update(lson, L, R, v);}if (R > mid) {update(rson, L, R, v);}push_up(rt);}pii query(int rt, int l, int r, int L, int R) {if (l >= L && r <= R) {return {minn[rt], maxn[rt]};}push_down(rt);pii ans = {0x3f3f3f3f, -0x3f3f3f3f};if (L <= mid) {pii res = query(lson, L, R);ans.first = min(ans.first, res.first);ans.second = max(ans.second, res.second);}if (R > mid) {pii res = query(rson, L, R);ans.first = min(ans.first, res.first);ans.second = max(ans.second, res.second);}return ans;}
}A, B;bool judge(int l1, int r1, int l2, int r2, int x, int y) {l1 = min(x, x + l1), r1 = max(x, x + r1);l2 = min(y, y + l2), r2 = max(y, y + r2);int L = max(l1, l2), R = min(r1, r2);return L <= R;
}int query(int rt, int l, int r, int min_a, int max_a, int min_b, int max_b, int x, int y) {if (l == r) {int cur_mina = min(min_a, A.minn[rt]), cur_maxa = max(max_a, A.maxn[rt]);int cur_minb = min(min_b, B.minn[rt]), cur_maxb = max(max_b, B.maxn[rt]);if (judge(cur_mina, cur_maxa, cur_minb, cur_maxb, x, y)) {return l;}return -1;}A.push_down(rt), B.push_down(rt);int cur_mina = min(min_a, A.minn[ls]), cur_maxa = max(max_a, A.maxn[ls]);int cur_minb = min(min_b, B.minn[ls]), cur_maxb = max(max_b, B.maxn[ls]);if (judge(cur_mina, cur_maxa, cur_minb, cur_maxb, x, y)) {return query(lson, min_a, max_a, min_b, max_b, x, y);}else {return query(rson, cur_mina, cur_maxa, cur_minb, cur_maxb, x, y);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &n, &m);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);sum[i] = sum[i - 1] + a[i];}A.build(1, 1, n);for (int i = 1; i <= n; i++) {scanf("%d", &b[i]);sum[i] = sum[i - 1] + b[i];}B.build(1, 1, n);for (int i = 1, op, x, y; i <= m; i++) {scanf("%d %d %d", &op, &x, &y);if (op == 1) {A.update(1, 1, n, x, n, -a[x]);a[x] = y;A.update(1, 1, n, x, n, a[x]);}else if (op == 2) {B.update(1, 1, n, x, n, -b[x]);b[x] = y;B.update(1, 1, n, x, n, b[x]);}else {if (x == y) {puts("0");continue;}printf("%d\n", query(1, 1, n, 0x3f3f3f3f, 0, 0x3f3f3f3f, 0, x, y));}}return 0;
}