牛客题解 | 【模板】二维差分

二维差分算法优化矩阵区域修改

最新推荐文章于 2025-09-16 09:11:21 发布

原创最新推荐文章于 2025-09-16 09:11:21 发布 · 1k 阅读

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题目

题目链接

解题思路

问题分析：
- 需要对矩阵进行多次子矩阵区域修改
- 直接修改会导致时间复杂度 $\mathcal{O}(n*m*q)$
- 可以使用二维差分数组优化到 $\mathcal{O}(n*m + q)$
二维差分数组优化：
- 对于区域 $(x 1, y 1, x 2, y 2)$ 加上 $k$ ，需要：
  - $d i ff [x 1] [y 1] + = k$
  - $d i ff [x 1] [y 2 + 1] - = k$
  - $d i ff [x 2 + 1] [y 1] - = k$
  - $d i ff [x 2 + 1] [y 2 + 1] + = k$
- 最后通过二维前缀和还原原矩阵
实现要点：
- 差分数组需要多开一行一列
- 注意数据范围，使用 $\text{long long}$ 类型
- 注意下标从 $1$ 开始

代码

#include <iostream>
#include <vector>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n, m, q;
    cin >> n >> m >> q;
    
    // 读入原矩阵
    vector<vector<long long>> a(n + 2, vector<long long>(m + 2, 0));
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            cin >> a[i][j];
        }
    }
    
    // 构建二维差分数组
    vector<vector<long long>> diff(n + 2, vector<long long>(m + 2, 0));
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            // 二维差分
            diff[i][j] += a[i][j];
            diff[i][j+1] -= a[i][j];
            diff[i+1][j] -= a[i][j];
            diff[i+1][j+1] += a[i][j];
        }
    }
    
    // 处理修改操作
    while(q--) {
        int x1, y1, x2, y2;
        long long k;
        cin >> x1 >> y1 >> x2 >> y2 >> k;
        // 二维差分修改
        diff[x1][y1] += k;
        diff[x1][y2+1] -= k;
        diff[x2+1][y1] -= k;
        diff[x2+1][y2+1] += k;
    }
    
    // 还原矩阵并输出
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            // 二维前缀和还原
            a[i][j] = a[i-1][j] + a[i][j-1] - a[i-1][j-1] + diff[i][j];
            cout << a[i][j] << (j == m ? '\n' : ' ');
        }
    }
    
    return 0;
}

import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        
        String[] line = br.readLine().split(" ");
        int n = Integer.parseInt(line[0]);
        int m = Integer.parseInt(line[1]);
        int q = Integer.parseInt(line[2]);
        
        // 读入原矩阵
        long[][] a = new long[n + 2][m + 2];
        for(int i = 1; i <= n; i++) {
            line = br.readLine().split(" ");
            for(int j = 1; j <= m; j++) {
                a[i][j] = Long.parseLong(line[j-1]);
            }
        }
        
        // 构建二维差分数组
        long[][] diff = new long[n + 2][m + 2];
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                diff[i][j] += a[i][j];
                diff[i][j+1] -= a[i][j];
                diff[i+1][j] -= a[i][j];
                diff[i+1][j+1] += a[i][j];
            }
        }
        
        // 处理修改操作
        while(q-- > 0) {
            line = br.readLine().split(" ");
            int x1 = Integer.parseInt(line[0]);
            int y1 = Integer.parseInt(line[1]);
            int x2 = Integer.parseInt(line[2]);
            int y2 = Integer.parseInt(line[3]);
            long k = Long.parseLong(line[4]);
            
            diff[x1][y1] += k;
            diff[x1][y2+1] -= k;
            diff[x2+1][y1] -= k;
            diff[x2+1][y2+1] += k;
        }
        
        // 还原矩阵并输出
        StringBuilder sb = new StringBuilder();
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                a[i][j] = a[i-1][j] + a[i][j-1] - a[i-1][j-1] + diff[i][j];
                sb.append(a[i][j]).append(j == m ? '\n' : ' ');
            }
        }
        out.print(sb);
        out.flush();
    }
}

def main():
    n, m, q = map(int, input().split())
    
    # 读入原矩阵
    a = [[0] * (m + 2) for _ in range(n + 2)]
    for i in range(1, n + 1):
        row = list(map(int, input().split()))
        for j in range(1, m + 1):
            a[i][j] = row[j-1]
    
    # 构建二维差分数组
    diff = [[0] * (m + 2) for _ in range(n + 2)]
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            diff[i][j] += a[i][j]
            diff[i][j+1] -= a[i][j]
            diff[i+1][j] -= a[i][j]
            diff[i+1][j+1] += a[i][j]
    
    # 处理修改操作
    for _ in range(q):
        x1, y1, x2, y2, k = map(int, input().split())
        diff[x1][y1] += k
        diff[x1][y2+1] -= k
        diff[x2+1][y1] -= k
        diff[x2+1][y2+1] += k
    
    # 还原矩阵并输出
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            a[i][j] = a[i-1][j] + a[i][j-1] - a[i-1][j-1] + diff[i][j]
            print(a[i][j], end='\n' if j == m else ' ')

if __name__ == "__main__":
    main()