米哈游笔试——求强势顶点的个数

2025/08/10 游戏客户端开发A卷

这里题目就不展开讲了，题目本身可以转换为查询数组区间上最大（小）值的个数。

害，做题时还没学线段树，所以只暴力了20%，现在给出线段树求解代码。

关于线段树可参考左神视频：

左程云--算法讲解110【扩展】线段树专题1-线段树原理和代码详解

#include <iostream>
#include <vector>
#include <unordered_map>using namespace std;
const static int N = 1e5 + 1;
vector<int> cArr = vector<int>(N, 0);
vector<vector<int>> tree = vector<vector<int>>(4 * N, vector<int>(2, 0));
pair<int, int> myUnion(int i, int j);
void build(int l, int r, int i);
pair<int, int> query(int jobl, int jobr, int l, int r, int i);
pair<int, int> myUnion(pair<int, int>& left, pair<int, int>& right);pair<int, int> myUnion(pair<int, int>& left, pair<int, int>& right) {if (left.first < right.first) {return left;}else if (right.first < left.first) {return right;}else {return { left.first, left.second + right.second };}
}void build(int l, int r, int i) {if (l == r) {tree[i][0] = cArr[l];tree[i][1] = 1;return;}int mid = l + ((r - l) >> 1);build(l, mid, i << 1);build(mid + 1, r, i << 1 | 1);auto ans = myUnion(i << 1, i << 1 | 1);tree[i][0] = ans.first;tree[i][1] = ans.second;
}pair<int, int> myUnion(int i, int j) {if (tree[i][0] < tree[j][0]) {return { tree[i][0], tree[i][1] };}else if (tree[i][0] > tree[j][0]) {return { tree[j][0], tree[j][1] };}else {return { tree[i][0], tree[i][1] + tree[j][1] };}
}pair<int, int> query(int jobl, int jobr, int l, int r, int i) {if (jobl <= l && r <= jobr) {return { tree[i][0], tree[i][1] };}int mid = l + ((r - l) >> 1);pair<int, int> left, right;if (jobl <= mid) {left = query(jobl, jobr, l, mid, i << 1);}if (jobr > mid) {right = query(jobl, jobr, mid + 1, r, i << 1 | 1);}if (left.second == 0) { return right; }if (right.second == 0) { return left; }return myUnion(left, right);
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n, q;	// n -> 数组大小  q -> 查询次数cin >> n >> q;//n = 5, q = 1;for (int i = 0; i < n; ++i) {cin >> cArr[i];	// ci = ai}//cArr = { 1, -2, 2, -2, -4 };int b;for (int i = 0; i < n; ++i) {cin >> b;cArr[i] = b - cArr[i];		// ci = bi - ai}build(0, n - 1, 1);while (q--) {int l, r;cin >> l >> r;l--; r--;auto ans = query(l, r, 0, n - 1, 1);cout << ans.second << endl;}return 0;
}