图像处理中的卷积操作实现

[题目链接](https://www.nowcoder.com/practice/798b474486af4528b9722feacc39aafb)

思路

本题要求对灰度图像矩阵进行二维卷积操作，使用零填充（zero padding）处理边界。

核心步骤：

零填充：设卷积核尺寸为 $k$ ，令 $p = \lfloor k/2 \rfloor$ 。在原始 $m \times n$ 图像的四周各填充 $p$ 层零，得到 $(m+2p) \times (n+2p)$ 的填充矩阵。

逐像素卷积：对输出矩阵的每个位置 $(i, j)$ ，从填充矩阵中取出以 $(i+p, j+p)$ 为中心的 $k \times k$ 区域，与卷积核逐元素相乘后求和：

$ $O(i,j) = \sum_{di=0}^{k-1} \sum_{dj=0}^{k-1} \text{padded}[i+di][j+dj] \cdot K[di][dj]$ $

格式化输出：每个结果用 Python 的 round(x, 2) 保留两位小数，再转字符串输出（尾部多余的零会被去掉）。

以样例为例：

图像为 $4 \times 4$ ，卷积核为 $3 \times 3$ ， $p = 1$ 。零填充后变为 $6 \times 6$ 。

对位置 $(0,0)$ ：取填充矩阵的 $[0:3, 0:3]$ 区域为 $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 5 & 6 \end{bmatrix}$ ，与卷积核逐元素相乘求和 = $0 \cdot 0 + (-1) \cdot 0 + 0 \cdot 0 + (-1) \cdot 0 + 5 \cdot 1 + (-1) \cdot 2 + 0 \cdot 0 + (-1) \cdot 5 + 0 \cdot 6 = -2.0$ ，与预期输出一致。

整个过程就是按定义直接模拟即可。

代码

C++
Java
Python3

#include <bits/stdc++.h>
using namespace std;

string fmt(double v) {
    v = round(v * 100.0) / 100.0;
    if (v == 0.0) v = 0.0;
    char buf[64];
    sprintf(buf, "%.2f", v);
    string s(buf);
    size_t dot = s.find('.');
    if (dot != string::npos) {
        size_t last = s.size() - 1;
        while (last > dot + 1 && s[last] == '0') last--;
        s = s.substr(0, last + 1);
    }
    return s;
}

int main() {
    int m, n;
    scanf("%d%d", &m, &n);
    vector<vector<double>> img(m, vector<double>(n));
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            scanf("%lf", &img[i][j]);
    int k;
    scanf("%d", &k);
    vector<vector<double>> ker(k, vector<double>(k));
    for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++)
            scanf("%lf", &ker[i][j]);

    int pad = k / 2;
    int pm = m + 2 * pad, pn = n + 2 * pad;
    vector<vector<double>> padded(pm, vector<double>(pn, 0.0));
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            padded[i + pad][j + pad] = img[i][j];

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            double val = 0.0;
            for (int di = 0; di < k; di++)
                for (int dj = 0; dj < k; dj++)
                    val += padded[i + di][j + dj] * ker[di][dj];
            if (j) printf(" ");
            printf("%s", fmt(val).c_str());
        }
        printf("\n");
    }
    return 0;
}

import java.util.*;

public class Main {
    static String fmt(double v) {
        v = Math.round(v * 100.0) / 100.0;
        if (v == 0.0) v = 0.0;
        String s = String.format("%.2f", v);
        int dot = s.indexOf('.');
        if (dot >= 0) {
            int last = s.length() - 1;
            while (last > dot + 1 && s.charAt(last) == '0') last--;
            s = s.substring(0, last + 1);
        }
        return s;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt(), n = sc.nextInt();
        double[][] img = new double[m][n];
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                img[i][j] = sc.nextDouble();
        int k = sc.nextInt();
        double[][] ker = new double[k][k];
        for (int i = 0; i < k; i++)
            for (int j = 0; j < k; j++)
                ker[i][j] = sc.nextDouble();

        int pad = k / 2;
        int pm = m + 2 * pad, pn = n + 2 * pad;
        double[][] padded = new double[pm][pn];
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                padded[i + pad][j + pad] = img[i][j];

        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                double val = 0.0;
                for (int di = 0; di < k; di++)
                    for (int dj = 0; dj < k; dj++)
                        val += padded[i + di][j + dj] * ker[di][dj];
                if (j > 0) sb.append(' ');
                sb.append(fmt(val));
            }
            sb.append('\n');
        }
        System.out.print(sb);
    }
}

import numpy as np

def main():
    m, n = map(int, input().split())
    image = []
    for _ in range(m):
        image.append(list(map(int, input().split())))
    k = int(input())
    kernel = []
    for _ in range(k):
        kernel.append(list(map(float, input().split())))

    img = np.array(image, dtype=float)
    ker = np.array(kernel, dtype=float)

    pad = k // 2
    padded = np.pad(img, pad, mode='constant', constant_values=0)

    out = np.zeros((m, n))
    for i in range(m):
        for j in range(n):
            region = padded[i:i+k, j:j+k]
            out[i][j] = np.sum(region * ker)

    for i in range(m):
        row = []
        for j in range(n):
            row.append(str(round(out[i][j], 2)))
        print(' '.join(row))

main()

复杂度分析

时间复杂度： $O(m \cdot n \cdot k^2)$ ，对输出矩阵的每个位置都需要遍历整个卷积核。
空间复杂度： $O((m+k) \cdot (n+k))$ ，存储零填充后的矩阵。