差分

能解决什么问题

快速的区间操作

算法思想

给定 a[1], a[2], ..., a[n]，构造差分数组 b[1], b[2], ..., b[n]，使得 a[i] = b[1] + b[2] + ... + b[i]。此时，如果将 a 数组 [l, r] 区间同时加上 c，等价于 b[l] += c，b[r + 1] -= c。

一维差分

#include <cstdio>

using namespace std;

const int N = 1e5 + 10;

int n, m;
int a[N], 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", &a[i]);
        insert(i, i, a[i]); // 相当于在 [i, i] 上加 a[i]
    }
    
    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]; // 求前缀和得到操作后的数组
        printf("%d ", b[i]);
    }
    
    return 0;
}

二维差分

#include <cstdio>

using namespace std;

const int N = 1010;

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

void insert(int x1, int y1, int x2, int y2, int c)
{
    b[x1][y1] += c;
    b[x1][y2 + 1] -= c;
    b[x2 + 1][y1] -= 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", &a[i][j]);
            insert(i, j, i, j, a[i][j]);
        }
        
    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;
}

三维差分

int b[A][B][C];

// 对于 insert 函数奇数个 1 是负号
int d[8][4] = {
    {0, 0, 0, 1};
    {0, 0, 1, -1};
    {0, 1, 0, -1};
    {0, 1, 1, 1};
    {1, 0, 0, -1};
    {1, 0, 1, 1};
    {1, 1, 0, 1};
    {1, 1, 1, -1};
};

// 是 1 的那位要变为二号点
void insert(int x1, int y1, int z1, int x2, int y2, int z2, int c)
{
    b[x1][y1][z1] += c;
    b[x1][y1][z2 + 1] -= c;
    b[x1][y2 + 1][z1] -= c;
    b[x1][y2 + 1][z2 + 1] += c;
    b[x2 + 1][y1][z1] -= c;
    b[x2 + 1][y1][z2 + 1] += c;
    b[x2 + 1][y2 + 1][z1] += c;
    b[x2 + 1][y2 + 1][z2 + 1] -= c;
}

void get_pre_sum(int i, int j, int k)
{
    for (int u = 1; u < 8; u++) {
        int x = i - d[u][0], y = j - d[u][1], z = k - d[u][2], t = -d[u][3]; // 求前缀和时除了 {0, 0, 0}，奇数个 1 是正号
        b[i][j][k] += b[x][y][z] * t;
    }
}