帮我给以下代码写注释void swap(int* a, int* b) { int tmp = *a; *a = *b, *b = tmp; } struct DisjointSetUnion { int *f, *size; int n, setCount; }; void initDSU(struct DisjointSetUnion* obj, int n) { obj->f = malloc(sizeof(int) * n); obj->size = malloc(sizeof(int) * n); obj->n = n; obj->setCount = n; for (int i = 0; i < n; i++) { obj->f[i] = i; obj->size[i] = 1; } } int find(struct DisjointSetUnion* obj, int x) { return obj->f[x] == x ? x : (obj->f[x] = find(obj, obj->f[x])); } int unionSet(struct DisjointSetUnion* obj, int x, int y) { int fx = find(obj, x), fy = find(obj, y); if (fx == fy) { return false; } if (obj->size[fx] < obj->size[fy]) { swap(&fx, &fy); } obj->size[fx] += obj->size[fy]; obj->f[fy] = fx; obj->setCount--; return true; } int connected(struct DisjointSetUnion* obj, int x, int y) { return find(obj, x) == find(obj, y); } struct Tuple { int x, y, z }; int cmp(const struct Tuple* a, const struct Tuple* b) { return a->z - b->z; } int minimumEffortPath(int** heights, int heightsSize, int* heightsColSize) { int m = heightsSize; int n = heightsColSize[0]; struct Tuple edges[n * m * 2]; int edgesSize = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { int id = i * n + j; if (i > 0) { edges[edgesSize].x = id - n; edges[edgesSize].y = id; edges[edgesSize++].z = fabs(heights[i][j] - heights[i - 1][j]); } if (j > 0) { edges[edgesSize].x = id - 1; edges[edgesSize].y = id; edges[edgesSize++].z = fabs(heights[i][j] - heights[i][j - 1]); } } } qsort(edges, edgesSize, sizeof(struct Tuple), cmp); struct DisjointSetUnion* uf = malloc(sizeof(struct DisjointSetUnion)); initDSU(uf, m * n); int ans = 0; for (int i = 0; i < edgesSize; i++) { unionSet(uf, edges[i].x, edges[i].y); if (connected(uf, 0, m * n - 1)) { ans = edges[i].z; break; } } return ans; }

/* 这个函数实现了两个数字交换的功能 输入：int* a, int* b, 分别是需要交换的两个数字的指针 输出：无 */ void swap(int* a, int* b) { int tmp = *a; *a = *b, *b = tmp; } /* 这个结构体实现了并查集的功能 包括初始化，查找根节点，合并两个集合，判断两个元素是否属于同一个集合 */ struct DisjointSetUnion { int *f, *size; // f数组表示每个元素的父节点，size数组表示每个集合的大小 int n, setCount; // n表示元素个数，setCount表示集合个数 }; /* 初始化并查集 输入：DisjointSetUnion* obj, 指向需要初始化的并查集的指针；int n，元素个数 输出：无 */ void initDSU(struct DisjointSetUnion* obj, int n) { obj->f = malloc(sizeof(int) * n); obj->size = malloc(sizeof(int) * n); obj->n = n; obj->setCount = n; for (int i = 0; i < n; i++) { obj->f[i] = i; obj->size[i] = 1; } } /* 查找元素所在集合的根节点 输入：DisjointSetUnion* obj, 指向并查集的指针；int x，需要查找的元素 输出：x所在集合的根节点 */ int find(struct DisjointSetUnion* obj, int x) { return obj->f[x] == x ? x : (obj->f[x] = find(obj, obj->f[x])); } /* 合并两个集合 输入：DisjointSetUnion* obj, 指向并查集的指针；int x, int y, 需要合并的两个元素 输出：合并是否成功，成功返回1，失败返回0 */ int unionSet(struct DisjointSetUnion* obj, int x, int y) { int fx = find(obj, x), fy = find(obj, y); if (fx == fy) { return false; } if (obj->size[fx] < obj->size[fy]) { swap(&fx, &fy); } obj->size[fx] += obj->size[fy]; obj->f[fy] = fx; obj->setCount--; return true; } /* 判断两个元素是否属于同一个集合 输入：DisjointSetUnion* obj, 指向并查集的指针；int x, int y，需要判断的两个元素 输出：是否属于同一个集合，是返回1，否返回0 */ int connected(struct DisjointSetUnion* obj, int x, int y) { return find(obj, x) == find(obj, y); } /* 这个结构体表示边 包括边的起点，终点和权值 */ struct Tuple { int x, y, z }; /* 比较边的权值，用于排序 输入：const struct Tuple* a, const struct Tuple* b，指向需要比较的两个边的指针 输出：两个边的权值之差 */ int cmp(const struct Tuple* a, const struct Tuple* b) { return a->z - b->z; } /* 这个函数实现了找到最小代价路径的功能 输入：int** heights，表示二维高度数组；int heightsSize，高度数组的行数；int* heightsColSize，高度数组的列数 输出：最小的代价值 */ int minimumEffortPath(int** heights, int heightsSize, int* heightsColSize) { int m = heightsSize; int n = heightsColSize[0]; struct Tuple edges[n * m * 2]; int edgesSize = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { int id = i * n + j; if (i > 0) { edges[edgesSize].x = id - n; edges[edgesSize].y = id; edges[edgesSize++].z = fabs(heights[i][j] - heights[i - 1][j]); } if (j > 0) { edges[edgesSize].x = id - 1; edges[edgesSize].y = id; edges[edgesSize++].z = fabs(heights[i][j] - heights[i][j - 1]); } } } qsort(edges, edgesSize, sizeof(struct Tuple), cmp); struct DisjointSetUnion* uf = malloc(sizeof(struct DisjointSetUnion)); initDSU(uf, m * n); int ans = 0; for (int i = 0; i < edgesSize; i++) { unionSet(uf, edges[i].x, edges[i].y); if (connected(uf, 0, m * n - 1)) { ans = edges[i].z; break; } } return ans; }