模板

• 差分是前缀和的逆运算
• 本质也是用空间换时间

一维差分

• 给出一个数组S构造出它的差分数组a使得S[i] = a[1] + … +a[i]
• 此时我们如果想在原数组S的某个区间上进行批量的数据修改，可以只在其差分数组中修改区间端点的值就行
• 这样时间复杂度就由o(n)降到了o(1)

构造差分数组：a[i] = S[i] - S[i - 1]
给区间[l, r]中的每个数加上c：a[l] += c, a[r + 1] -= c
    
或者构造一个insert方法：
public static void insert(int l, int r, int c) {
    a[l] += c;
    a[r + 1] -= c;
}
这个方法还可以巧妙地构造差分数组：
insert(i, i, S[i]);

二维差分

构造差分矩阵：a[i][j] = S[i][j] - S[i - 1][j] - S[i][j - 1] + S[i - 1][j - 1]
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：
S[x1, y1] += c, S[x2 + 1, y1] -= c, S[x1, y2 + 1] -= c, S[x2 + 1, y2 + 1] += c
    
或者构造一个insert方法：
public static void insert(int x1, int y1, int x2, int y2, int c) {
    S[x1, y1] += c;
    S[x2 + 1, y1] -= c;
    S[x1, y2 - 1] -= c;
    S[x2 + 1, y2 += 1] += c;
}
这个方法还可以巧妙地构造差分数组：
insert(i, j, i, j, S[i][j]);

练习

01 差分

• 题目

• 题解

import java.io.*;

public class Main{
    public static final int N = 100010;
    public static int[] S = new int[N];
    public static int[] a = new int[N];
    public static int n;
    public static int m;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        String[] str1 = br.readLine().split(" ");
        n = Integer.parseInt(str1[0]);
        m = Integer.parseInt(str1[1]);
        String[] str2 = br.readLine().split(" ");
        for (int i = 1; i <= n; i++) {
            S[i] = Integer.parseInt(str2[i - 1]);
        }

        //很巧妙的方法构造差分数组
        for (int i = 1; i <= n; i++) {
            insert(i, i, S[i]);
        }

        while (m-- > 0) {
            String[] str3 = br.readLine().split(" ");
            int l = Integer.parseInt(str3[0]);
            int r = Integer.parseInt(str3[1]);
            int c = Integer.parseInt(str3[2]);
            insert(l, r, c);
        }
        for (int i = 1; i <= n; i++) {
            S[i] = S[i - 1] + a[i];
        }
        for (int i = 1; i <= n; i++) {
            pw.print(S[i] + " ");
        }
        
        pw.close();
        br.close();
    }

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

02 差分矩阵

• 题目

• 题解

import java.io.*;

public class Main {
    public static final int N = 1010;
    public static int[][] a = new int[N][N];
    public static int[][] S = new int[N][N];
    public static int n, m, q;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        String[] str1 = br.readLine().split(" ");
        n = Integer.parseInt(str1[0]);
        m = Integer.parseInt(str1[1]);
        q = Integer.parseInt(str1[2]);
        
        //将初始化原矩阵和对应的差分矩阵结合起来
        for (int i = 1; i <= n; i++) {
            String[] str2 = br.readLine().split(" ");
            for (int j = 1; j <= m; j++) {
                S[i][j] = Integer.parseInt(str2[j - 1]);
                insert(i, j, i, j, S[i][j]);
            }
        }

        while (q-- > 0) {
            String[] str3 = br.readLine().split(" ");
            int x1 = Integer.parseInt(str3[0]);
            int y1 = Integer.parseInt(str3[1]);
            int x2 = Integer.parseInt(str3[2]);
            int y2 = Integer.parseInt(str3[3]);
            int c = Integer.parseInt(str3[4]);
            insert(x1, y1, x2, y2, c);
        }

        //将生成操作后的矩阵和输出矩阵结合在一起
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                S[i][j] = S[i - 1][j] + S[i][j - 1] - S[i - 1][j - 1] + a[i][j];
                pw.print(S[i][j] + " ");
            }
            pw.println();
        }
        pw.close();
        br.close();
    }

    public static void insert(int x1, int y1, int x2, int y2, int c) {
        a[x1][y1] += c;
        a[x2 + 1][y1] -= c;
        a[x1][y2 + 1] -= c;
        a[x2 + 1][y2 + 1] += c;
    }
}