【基础算法】差分

差分

// 每日一题
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 200010;
int n;
int a[N], s[N];
int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		s[i] = s[i - 1] + a[i]; // 前缀和
	}
	LL res = 0;
	for (int i = 1; i <= n - 1; i++)
	{
		res = res + (LL)a[i] * (s[n] - s[i]);
	}
	cout << res << endl;
	return 0;
}

什么是差分？

差分就是一个序列中后一项减前一项的过程。

差分数组就是一个序列中后一项减前一项的值组成的一个新的序列。

例如

有一个序列为：$[2,1,4,6,3,2]$ 那么这个序列的差分数组为：$[2,-1,3,2,-3,-1]$

差分可以解决什么问题？

• 区间修改（将 $O(n)$ 时间的修改区间变为 $O(1)$ 时间的修改端点）

例题：一维差分

暴力：

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, m;
int a[N];
int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i++) 
	{
		cin >> a[i];
	}
	while (m --)
	{
		int l, r, c;
		cin >> l >> r >> c;
		for (int i = l; i <= r; i ++)
			a[i] += c;
	}
	for (int i = 1; i <= n; i++) cout << a[i] << ' ';
	return 0;
}

差分：

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, m;
int a[N], b[N];
int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i++) 
	{
		cin >> a[i];
		b[i] = a[i] - a[i - 1]; // 初始化差分数组
	}
	while (m --)
	{
		int l, r, c;
		cin >> l >> r >> c;
		b[l] += c;
		b[r + 1] -= c;
	}
	for (int i = 1; i <= n; i++) b[i] += b[i - 1]; // 前缀和
	
	for (int i = 1; i <= n; i++) cout << b[i] << ' ';
	return 0;
}

例题：二维差分

暴力：

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N];
int main()
{
	cin >> n >> m >> q;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			cin >> a[i][j];
	while (q --)
	{
		int x1, y1, x2, y2, c;
		cin >> x1 >> y1 >> x2 >> y2 >> c;
		for (int i = x1; i <= x2; i++)
			for (int j = y1; j <= y2; j++)
				a[i][j] += c;
	}
	
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
			cout << a[i][j] << ' ';
		cout << endl;
	}
	return 0;
}

差分：

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N], b[N][N];
void modify(int x1, int y1, int x2, int y2, int c)
{
	b[x1][y1] += c;
	b[x2 + 1][y1] -= c;
	b[x1][y2 + 1] -= c;
	b[x2 + 1][y2 + 1] += c;
}
int main()
{
	cin >> n >> m >> q;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
		{
			cin >> a[i][j];
			//b[i][j] = a[i][j] - a[i - 1][j] - a[i][j - 1] + a[i - 1][j - 1]; // 初始化差分数组
     		modify(i, j, i, j, a[i][j]); // 初始化差分数组       
		}
	while (q--)
	{
		int x1, y1, x2, y2, c;
		cin >> x1 >> y1 >> x2 >> y2 >> c;
		modify(x1, y1, x2, y2, c);
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1]; // 求前缀和
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
			cout << b[i][j] << ' ';
		cout << endl;
	}
	return 0;
}