区间问题
ST表(静态区间查找)
ST表是利用倍增思想来缩短时间的,数组 f[i][j] 的含义,从第 i 个元素开始，向后连续 2^j 个元素组成的区间内的最值（最大值或最小值，需提前确定）。
\[f[i][j] = \max\left(f[i][j-1],\ f\left[i + 2^{j-1}\right][j-1]\right)
\]
void solved()
{int n, t;cin >> n >> t;for (int i = 1; i <= n; i++)cin >> stmax[i][0];for (int j = 1; (1 << j) <= n; j++){for (int i = 1; i + (1 << j) - 1 <= n; i++){stmax[i][j] = max(stmax[i][j - 1], stmax[i + (1 << (j - 1))][j - 1]);}}while (t--){int l, r;cin >> l >> r;int s = log2(r - l + 1);cout << max(stmax[l][s], stmax[r + 1 - (1 << s)][s]) << endl;}
}
树状数组
单修区查
int lowbit(int x)
{return x & -x;
}int query(int x)
{int sum = 0;for(int i = x; i; i -=lowbit(i)){sum += c[i];}return sum;
}void update(int x, int v)
{for(int i = x; i <= n; i += lowbit(i)){c[i] += v;}
}
void solved()
{cin >> n >> t;for (int i = 1; i <= n; i++){cin >> arr[i];update(i, arr[i]);}while(t --){int op, x, y, k;cin >> op;if(op == 1){cin >> x >> k;update(x, k);}else {cin >> x >> y;cout << query(y) - query(x - 1) << endl;}}
}
区修单查（差分）
把t[i]构建成差分数组，query(x)时求(1, x)前缀和就是点x的值
//构建差分数组
update(i, x - last);
//修改区间
update(x, k);
update(y + 1, -k);
区修区查
int lowbit(int x)
{return x & -x;
}void update(int x, LL c)
{for (int i = x; i <= n; i += lowbit(i)){b1[i] += c, b2[i] += x * c;}
}LL query(int x)
{LL res = 0;for (int i = x; i; i -= lowbit(i)){res += (x + 1) * b1[i] - b2[i];}return res;
}题目链接二维模板题
二维树状数组模板（区修区查），公式推导过程太复杂，记下就好(其实就是不会)_ 根二维差分数组类似
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;const int N = 2050;int n, m ,q;LL a[N][N], b[N][N], c[N][N], d[N][N];int lowbit(int x){return x & -x;
}void add(int x, int y, LL w){for(int i = x; i <= n; i += lowbit(i)){for(int j = y; j <= m; j += lowbit(j)){a[i][j] += w;b[i][j] += (x - 1) * w;c[i][j] += (y - 1) * w;d[i][j] += (x - 1) * (y - 1) * w;}}
}LL query(int x, int y){LL res = 0;for(int i = x; i > 0; i -= lowbit(i)){for(int j = y; j > 0; j -= lowbit(j)){res += x * y * a[i][j] - y * b[i][j]- x * c[i][j]+ d[i][j];}}return res;
}int main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cin >> n >> m >> q;while (q --) {int op, x1, y1, x2, y2;cin >> op >> x1 >> y1 >> x2 >> y2;if (op == 1) {LL v;cin >> v;add(x1, y1, v);add(x1, y2 + 1, -v);add(x2 + 1, y1, -v);add(x2 + 1, y2 + 1, v);} else {cout << query(x2, y2) - query(x1 - 1, y2) - query(x2, y1 - 1) + query(x1 - 1, y1 - 1)<< '\n';}}
}
线段树
单修区查
struct node
{int l, r, sum;
} tree[4 * N];void build(int i, int l, int r)
{tree[i].l = l;tree[i].r = r;if (l == r){tree[i].sum = arr[l];return;}int mid = (l + r) >> 1;build(i * 2, l, mid);build(i * 2 + 1, mid + 1, r);tree[i].sum = tree[i * 2].sum + tree[i * 2 + 1].sum;
}void update(int i, int x, int k)
{if (tree[i].l == tree[i].r){tree[i].sum += k;return;}if (x <= tree[i * 2].r)update(i * 2, x, k);elseupdate(i * 2 + 1, x, k);tree[i].sum = tree[i * 2].sum + tree[i * 2 + 1].sum;
}int query(int i, int l, int r)
{if (tree[i].l >= l && tree[i].r <= r){return tree[i].sum;}if (tree[i].l > r || tree[i].r < l)return 0;int res = 0;if (tree[i * 2].r >= l)res += query(i * 2, l, r);if (tree[i * 2 + 1].l <= r)res += query(i * 2 + 1, l, r);return res;
}
区修区查(lazy 延迟更新)
线段树的区间更新（如给 [L, R] 内所有元素加 delta），如果不使用懒标记，需要递归遍历所有与 [L, R] 重叠的叶子节点并逐个更新，在最坏情况下（如更新整个数组）时间复杂度为 O (n)，效率极低。
当更新的区间完全覆盖某个节点的区间时，不立即更新该节点的子节点，而是给该节点打上一个 “标记”，记录需要更新的内容。只有当后续操作（如查询、再次更新）需要访问该节点的子节点时，才将标记 “下推” 到子节点，确保子节点的状态正确。
struct node
{int l, r, sum, lazy;
} tree[4 * N];void build(int i, int l, int r)
{tree[i].l = l;tree[i].r = r;tree[i].lazy = 0;tree[i].sum = 0;if (l == r){tree[i].sum = arr[l];return;}int mid = (l + r) >> 1;build(i * 2, l, mid);build(i * 2 + 1, mid + 1, r);tree[i].sum = tree[i * 2].sum + tree[i * 2 + 1].sum;
}void pushdown(int i)
{if (tree[i].lazy){tree[i * 2].lazy += tree[i].lazy;tree[i * 2 + 1].lazy += tree[i].lazy;int mid = (tree[i].l + tree[i].r) >> 1;tree[i * 2].sum += tree[i].lazy * (mid - tree[i * 2].l + 1);tree[i * 2 + 1].sum += tree[i].lazy * (tree[i * 2 + 1].r - mid + 1);tree[i].lazy = 0;}
}
void update(int i, int l, int r, int k)
{if (tree[i].l >= l && tree[i].r <= r){tree[i].sum += k * (tree[i].r - tree[i].l + 1);tree[i].lazy += k;return;}pushdown(i);if (tree[i * 2].r >= l)update(i * 2, l, r, k);if (tree[i * 2 + 1].l <= r)update(i * 2 + 1, l, r, k);tree[i].sum = tree[i * 2].sum + tree[i * 2 + 1].sum;
}int query(int i, int l, int r)
{if (tree[i].l >= l && tree[i].r <= r){return tree[i].sum;}if (tree[i].l > r || tree[i].r < l)return 0;pushdown(i);int res = 0;if (tree[i * 2].r >= l)res += query(i * 2, l, r);if (tree[i * 2 + 1].l <= r)res += query(i * 2 + 1, l, r);return res;
}
