第一章：6.差分+前缀和（一个区域整体添加一个数）

1.【模板】差分

 

//#include<iostream>
//
//using namespace std;
//
//#include<iostream>
//
//using namespace std;
//
//typedef long long LL;
//
//const int N = 1e5 + 10;
//int n, m;
//LL a[N],f[N];
//
//int main()
//{
//	cin >> n >> m;
//	for (int i = 1; i <= n; i++)
//	{
//		cin >> a[i];
//		f[i] = a[i] - a[i - 1];
//	}
//
//	while (m--)
//	{
//		int l, r, k; cin >> l >> r >> k;
//		f[l] += k, f[r + 1] -= k;
//	}
//
//	for (int i = 1; i <= n; i++)
//	{
//		a[i] = a[i - 1] + f[i];
//		cout << a[i] << " ";
//	}
//	return 0;
//}
#include<iostream>

using namespace std;

#include<iostream>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;
int n, m;
LL a[N], f[N];

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		f[i] += a[i];
		f[i + 1] -= a[i];
	}

	while (m--)
	{
		int l, r, k; cin >> l >> r >> k;
		f[l] += k, f[r + 1] -= k;
	}

	for (int i = 1; i <= n; i++)
	{
		f[i] = f[i] + f[i - 1];
		cout << f[i]<<"  ";
	}
	return 0;
}

2.P3406 海底高铁 - 洛谷

 

#include<iostream>

using namespace std;

typedef long long LL;

int n, m;//n个收费点,个站
const int N = 1e3 + 10;
int a[N];
LL f[N];
LL sum = 0;

int main()
{
	cin >> n >> m;

	int x; cin >> x;
	for (int i = 2; i <= m; i++)
	{
		int  y; cin >> y;
		if (x < y)
		{
			f[x]++;
			f[y]--;
		}
		else if(x > y)
		{
			f[y]++;
			f[x]--;
		}
		x = y;
	}

	for (int i = 1; i < n; i++)
	{
		int ai, bi, ci; cin >> ai >> bi >> ci;
		f[i] = f[i] + f[i - 1];
		sum += min(f[i] * ai, ci + f[i] * bi);
	}
	cout << sum << endl;
	return 0;
}

3.【模板】二维差分

 

 

#include<iostream>

using namespace std;

typedef long long LL;

int n, m, q;
const int N = 1e3 + 10;
LL f[N][N];

void insert(int x1, int y1, int x2, int y2, LL k)
{
	f[x1][y1] += k; f[x2 + 1][y1] -= k; f[x1][y2 + 1] -= k; f[x2 + 1][y2 + 1] += k;
}

int main()
{
	cin >> n >> m >> q;
	//预处理二维数组
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			LL x; cin >> x;
			//以（i,j)为左上角，(i,j)为右下脚，进行插入
			insert(i, j, i, j, x);
		}
	}
	while (q--)
	{
		int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
		LL v; cin >> v;
		insert(x1, y1, x2, y2, v);
	}
	//整理合成最终矩阵，利用二维数组的前缀和
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			f[i][j] = f[i - 1][j] + f[i][j - 1] - f[i - 1][j - 1] + f[i][j];
			cout << f[i][j] << " ";
		}
		cout << endl;
	}
}

利用二维查分填表

最后利用前缀和完善数组 

4.P3397 地毯 - 洛谷（第三题的模版题）

 

#include<iostream>

using namespace std;

typedef long long LL;

int n, m;
const int N = 1e3 + 10;
LL f[N][N];

void insert(int x1, int y1, int x2, int y2, LL k)
{
	f[x1][y1] += k; f[x2 + 1][y1] -= k; f[x1][y2 + 1] -= k; f[x2 + 1][y2 + 1] += k;
}

int main()
{
	cin >> n >> m;
	//预处理二维数组
	while (m--)
	{
		int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
		insert(x1, y1, x2, y2, 1);
	}
	//整理合成最终矩阵，利用二维数组的前缀和
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			f[i][j] = f[i - 1][j] + f[i][j - 1] - f[i - 1][j - 1] + f[i][j];
			cout << f[i][j] << " ";
		}
		cout << endl;
	}
}