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最新推荐文章于 2025-12-12 16:37:00 发布

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P2880 [USACO07JAN] Balanced Lineup G - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

这道题可以用 ST 表来做，存两个然后分别求最大最小值，之后查找相减即可。

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int logn = 21;
const int maxn = 50010;
int f[maxn][logn + 1], Logn[maxn + 1]; // 这里存数组的第二维下标
int fa[maxn][logn + 1];
int fi[maxn][logn + 1];

inline int read()
{
	char c = getchar();
	int x = 0, f = 1;
	while (c < '0' || c > '9')
	{
		if (c == '-')
			f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
	{
		x = x * 10 + c - '0';
		c = getchar();
	}
	return f * x;
}

void pre()
{
	Logn[1] = 0;
	Logn[2] = 1;
	for (int i = 3; i < maxn; i++)
	{
		Logn[i] = Logn[i / 2] + 1;
	}
}

int main()
{
	int n = read(), q = read();
	pre();
	for (int i = 1; i <= n; i++)
	{
		f[i][0] = read();
		fi[i][0] = f[i][0];
	}
	for (int j = 1; j <= logn; j++)
	{
		for (int i = 1; i + (1 << j) - 1 <= n; i++)
		{
			f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
		}
	}
	for (int j = 1; j <= logn; j++)
	{
		for (int i = 1; i + (1 << j) - 1 <= n; i++)
		{
			fi[i][j] = min(fi[i][j - 1], fi[i + (1 << (j - 1))][j - 1]);
		}
	}
	for (int i = 1; i <= q; i++)
	{
		int a = read(), b = read();
		int s = Logn[b - a + 1];
		printf("%d\n", max(f[a][s], f[b - (1 << s) + 1][s]) - min(fi[a][s], fi[b - (1 << s) + 1][s]));
	}
	
	return 0;
}