问题描述

依然是RMQ问题。过程需要修改某些点的数据。

使用线段树解决。

单点修改的复杂度是O(logn)，即树深，只需要修改这个点的祖先节点。

区间查询的复杂度也是O(logn)，因为除了第一步可能一分为二外，其他查询若有分解则分解后必然有一个区间是不需要再分解的。

#include <stdio.h>
#include <algorithm>typedef
struct _seg_tree_ {int left, right;int val;_seg_tree_ *lson = NULL, *rson = NULL;_seg_tree_(int left_idx, int right_idx, int value): left(left_idx), right(right_idx), val(value) {}
} seg_tree, *pseg_tree;pseg_tree construct_seg_tree(int left, int right) {if (left == right) {int val;scanf("%d", &val);return new seg_tree(left, left, val);}int mid = left + (right - left) / 2;pseg_tree lson = construct_seg_tree(left, mid);pseg_tree rson = construct_seg_tree(mid + 1, right);int val = std::min(lson->val, rson->val);pseg_tree ans = new seg_tree(left, right, val);ans->lson = lson;ans->rson = rson;return ans;
}int query_seg_tree(pseg_tree proot, int left, int right) {if (left == proot->left && right == proot->right) {return proot->val;}int mid = proot->left + (proot->right - proot->left) / 2;if (left > mid) {return query_seg_tree(proot->rson, left, right);}if (right <= mid) {return query_seg_tree(proot->lson, left, right);}return std::min(query_seg_tree(proot->lson, left, mid),query_seg_tree(proot->rson, mid + 1, right));
}void modify_seg_tree(pseg_tree proot, int left, int val) {if (left == proot->left && left == proot->right) {proot->val = val;return;}int mid = proot->left + (proot->right - proot->left) / 2;if (left > mid) {modify_seg_tree(proot->rson, left, val);} else {modify_seg_tree(proot->lson, left, val);}proot->val = std::min(proot->lson->val, proot->rson->val);
}int main(){int n, q;scanf("%d", &n);pseg_tree proot = construct_seg_tree(1, n);scanf("%d", &q);int op, left, right;while (q--) {scanf("%d%d%d", &op, &left, &right);if (op == 0) {int ans = query_seg_tree(proot, left, right);printf("%d\n", ans);} else {modify_seg_tree(proot, left, right);}}return 0;
}