差分算法总结

差分是前缀和的逆运算

一维差分

对于a1,a2,…,an,构造b1,b2,…,bn,使得ai = b1 + b+ … + bi。此时,b数组成为a数组的差分,a数组称为b数组的前缀和。

题目链接:

https://www.acwing.com/problem/content/799/

代码模版:

#include <iostream>

using namespace std;

const int N = 100010;

int n, m, x;
int b[N];

void insert(int l, int r, int c)
{
    b[l] += c;
    b[r + 1] -= c;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &x);
        insert(i, i, x);
    }

    while (m--)
    {
        int l, r, c;
        scanf("%d%d%d", &l, &r, &c);
        insert(l, r, c);
    }

    for (int i = 1; i <= n; i++)
    {
        b[i] += b[i - 1];
        printf("%d ", b[i]);
    }

    return 0;
}

 

二维差分

原矩阵为A = (aij)n*m,差分矩阵为= (bij)n*m,使得矩阵A是差分矩阵B的前缀和。

初始化:令矩阵A = O,那么矩阵= O。然后把矩阵A中的每个元素依次插入。

题目链接:

https://www.acwing.com/problem/content/800/

代码模版:

#include <iostream>

using namespace std;

const int N = 1010;

int n, m, q, x;
int b[N][N];

void insert(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()
{
    scanf("%d%d%d", &n, &m, &q);

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            scanf("%d", &x);
            insert(i, j, i, j, x);
        }

    while (q--)
    {
        int x1, y1, x2, y2, c;
        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &c);
        insert(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];
            printf("%d ", b[i][j]);
        }
        puts("");
    }

    return 0;
}

 

posted @ 2023-11-27 23:33  ykycode  阅读(28)  评论(0编辑  收藏  举报