5-差分

算法思想

1. 计算出差分
2. 通过差分求解区间问题

一维差分

a 为普通数组，b 为 a 的差分数组，
则 a 与 b 的关系如下：

b[1] = a[1]  
b[i] = a[i] - a[i - 1]

数组 a 的 [l, r] 区间上的数都加上 c，  
等价于 
b[l] += c，
b[r + 1] -= c

二维差分

a 为普通矩阵（二维数组），b 为 a 的差分矩阵， 则 a 与 b 的关系如下：

a[1][1] = b[1][1]
a[i][j] = a[i - 1][j] + a[i][j - 1] - a[i][j]+ b[i][j]
（此关系需要补0才能发现）

b[i][j] + c => 
a[x][y] + c, i <= x <= n, j <= y <= m

给以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵中的所有元素加上 c，等价于
b[x1, y1] += c, 
b[x2 + 1, y1] -= c,
b[x1, y2 + 1] -= c, 
b[x2 + 1, y2 + 1] += c

算法模板

一维差分

const int N = 100010;
int n, m;
//初始化数组a及其差分数组b；
int a[N], b[N];

int main(void)
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);

    // 更新a的差分数组b
    for(int i = 1; i <= n; i++) insert(i, i, a[i]);

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

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

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

二维差分

const int N = 1010;
int n, m, q;
// 初始化矩阵a及其差分矩阵b
int a[N][N], b[N][N];

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", &a[i][j]);

    // 更新差分矩阵b
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            insert(i, j, i, j, a[i][j]);

    while (q -- )
    {
        int x1, y1, x2, y2, c;
        cin >> 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];

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

    return 0;
}

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;
}