请用C代码或C++代码，实现两种骨架提取算法：顺序细化和北、南、东、西提取算法。

顺序细化算法的C++代码实现如下：

#include <iostream>
#include <vector>

using namespace std;

// 顺序细化算法
void seqThin(vector<vector<int>>& image)
{
    int height = image.size();
    int width = image[0].size();

    // 循环次数
    int count = 0;

    while (true) {
        ++count;

        // 标记需要删除的点
        vector<pair<int, int>> toDelete;

        // 从上往下扫描
        for (int i = 1; i < height - 1; ++i) {
            for (int j = 1; j < width - 1; ++j) {
                if (image[i][j] == 0) {
                    continue;
                }

                int A = (image[i-1][j] == 0) ? 0 : 1;
                int B = (image[i-1][j+1] == 0) ? 0 : 1;
                int C = (image[i][j+1] == 0) ? 0 : 1;
                int D = (image[i+1][j+1] == 0) ? 0 : 1;
                int E = (image[i+1][j] == 0) ? 0 : 1;
                int F = (image[i+1][j-1] == 0) ? 0 : 1;
                int G = (image[i][j-1] == 0) ? 0 : 1;
                int H = (image[i-1][j-1] == 0) ? 0 : 1;

                int p1 = A + B + C + D + E + F + G + H;
                int p2 = (!A && B) + (!B && C) + (!C && D) + (!D && E) +
                         (!E && F) + (!F && G) + (!G && H) + (!H && A);
                int p3 = (A || B || !C) && (D || E || !C) &&
                         (!A || !D || !G) && (!B || !E || !G);

                if (p1 >= 2 && p1 <= 6 && p2 == 1 && p3) {
                    toDelete.push_back(make_pair(i, j));
                }
            }
        }

        // 删除标记的点
        for (auto& p : toDelete) {
            image[p.first][p.second] = 0;
        }

        // 标记需要删除的点
        toDelete.clear();

        // 从下往上扫描
        for (int i = height - 2; i >= 1; --i) {
            for (int j = width - 2; j >= 1; --j) {
                if (image[i][j] == 0) {
                    continue;
                }

                int A = (image[i-1][j] == 0) ? 0 : 1;
                int B = (image[i-1][j+1] == 0) ? 0 : 1;
                int C = (image[i][j+1] == 0) ? 0 : 1;
                int D = (image[i+1][j+1] == 0) ? 0 : 1;
                int E = (image[i+1][j] == 0) ? 0 : 1;
                int F = (image[i+1][j-1] == 0) ? 0 : 1;
                int G = (image[i][j-1] == 0) ? 0 : 1;
                int H = (image[i-1][j-1] == 0) ? 0 : 1;

                int p1 = A + B + C + D + E + F + G + H;
                int p2 = (!A && B) + (!B && C) + (!C && D) + (!D && E) +
                         (!E && F) + (!F && G) + (!G && H) + (!H && A);
                int p3 = (A || B || !C) && (D || E || !C) &&
                         (!A || !D || !G) && (!B || !E || !G);

                if (p1 >= 2 && p1 <= 6 && p2 == 1 && p3) {
                    toDelete.push_back(make_pair(i, j));
                }
            }
        }

        // 删除标记的点
        for (auto& p : toDelete) {
            image[p.first][p.second] = 0;
        }

        // 如果没有点被删除，退出循环
        if (toDelete.empty()) {
            break;
        }
    }

    cout << "循环次数：" << count << endl;
}

int main()
{
    vector<vector<int>> image = {
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 0, 1, 1, 0, 0, 0, 0, 0},
        {0, 0, 0, 1, 1, 0, 0, 0, 0, 0},
        {0, 0, 0, 0, 0, 1, 0, 0, 0, 0},
        {0, 0, 0, 0, 0, 1, 1, 0, 0, 0},
        {0, 0, 0, 0, 0, 0, 1, 0, 0, 0},
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
    };

    seqThin(image);

    for (auto& row : image) {
        for (auto& pixel : row) {
            cout << pixel << " ";
        }
        cout << endl;
    }

    return 0;
}

北、南、东、西提取算法的C代码实现如下：

#include <stdio.h>
#include <stdlib.h>

// 北、南、东、西提取算法
void bnsThin(int** image, int height, int width)
{
    // 循环次数
    int count = 0;

    while (1) {
        ++count;

        // 标记需要删除的点
        int toDelete[height * width][2];
        int numToDelete = 0;

        // 从上往下扫描
        for (int i = 1; i < height - 1; ++i) {
            for (int j = 1; j < width - 1; ++j) {
                if (image[i][j] == 0) {
                    continue;
                }

                int A = (image[i-1][j] == 0) ? 0 : 1;
                int B = (image[i-1][j+1] == 0) ? 0 : 1;
                int C = (image[i][j+1] == 0) ? 0 : 1;
                int D = (image[i+1][j+1] == 0) ? 0 : 1;
                int E = (image[i+1][j] == 0) ? 0 : 1;
                int F = (image[i+1][j-1] == 0) ? 0 : 1;
                int G = (image[i][j-1] == 0) ? 0 : 1;
                int H = (image[i-1][j-1] == 0) ? 0 : 1;

                int p1 = A + B + C + D + E + F + G + H;
                int p2 = (!A && B) + (!B && C) + (!C && D) + (!D && E) +
                         (!E && F) + (!F && G) + (!G && H) + (!H && A);
                int p3 = (A || B || !C) && (D || E || !C) &&
                         (!A || !D || !G) && (!B || !E || !G);

                if (p1 >= 2 && p1 <= 6 && p2 == 1 && p3) {
                    toDelete[numToDelete][0] = i;
                    toDelete[numToDelete][1] = j;
                    ++numToDelete;
                }
            }
        }

        // 删除标记的点
        for (int i = 0; i < numToDelete; ++i) {
            image[toDelete[i][0]][toDelete[i][1]] = 0;
        }

        // 标记需要删除的点
        numToDelete = 0;

        // 从下往上扫描
        for (int i = height - 2; i >= 1; --i) {
            for (int j = width - 2; j >= 1; --j) {
                if (image[i][j] == 0) {
                    continue;
                }

                int A = (image[i-1][j] == 0) ? 0 : 1;
                int B = (image[i-1][j+1] == 0) ? 0 : 1;
                int C = (image[i][j+1] == 0) ? 0 : 1;
                int D = (image[i+1][j+1] == 0) ? 0 : 1;
                int E = (image[i+1][j] == 0) ? 0 : 1;
                int F = (image[i+1][j-1] == 0) ? 0 : 1;
                int G = (image[i][j-1] == 0) ? 0 : 1;
                int H = (image[i-1][j-1] == 0) ? 0 : 1;

                int p1 = A + B + C + D + E + F + G + H;
                int p2 = (!A && B) + (!B && C) + (!C && D) + (!D && E) +
                         (!E && F) + (!F && G) + (!G && H) + (!H && A);
                int p3 = (A || B || !C) && (D || E || !C) &&
                         (!A || !D || !G) && (!B || !E || !G);

                if (p1 >= 2 && p1 <= 6 && p2 == 1 && p3) {
                    toDelete[numToDelete][0] = i;
                    toDelete[numToDelete][1] = j;
                    ++numToDelete;
                }
            }
        }

        // 删除标记的点
        for (int i = 0; i < numToDelete; ++i) {
            image[toDelete[i][0]][toDelete[i][1]] = 0;
        }

        // 如果没有点被删除，退出循环
        if (numToDelete == 0) {
            break;
        }
    }

    printf("循环次数：%d\n", count);
}

int main()
{
    int height = 10;
    int width = 10;

    int** image = (int**)malloc(height * sizeof(int*));
    for (int i = 0; i < height; ++i) {
        image[i] = (int*)malloc(width * sizeof(int));
        for (int j = 0; j < width; ++j) {
            image[i][j] = 0;
        }
    }

    image[2][3] = 1;
    image[2][4] = 1;
    image[3][3] = 1;
    image[3][4] = 1;
    image[4][5] = 1;
    image[5][5] = 1;
    image[6][4] = 1;
    image[5][3] = 1;

    bnsThin(image, height, width);

    for (int i = 0; i < height; ++i) {
        for (int j = 0; j < width; ++j) {
            printf("%d ", image[i][j]);
        }
        printf("\n");
    }

    for (int i = 0; i < height; ++i) {
        free(image[i]);
    }
    free(image);

    return 0;
}

注意：这里提供的代码仅作为示例，实际应用中需要根据具体情况进行修改。