Question

我们给出了不同的数字对。现在我们必须创建一个链，如果pair1的任何元素等于pair2的任何元素，那么它们属于同一个集合。

实施例： -

给出对：

(0,1)
(2,3)
(5,4)
(3,5)
(7,6)
(8,7)

所以这一切都是

{0,1}, {2,3,4,5}, {6,7,8}

我们如何实现这一目标。我对此并不在意。任何帮助将不胜感激。

解决方案 C代码 #include

#define SIZE 100100

int visited[SIZE];



struct adjListNode{
    int dest;
    struct adjListNode* next;
};

struct adjList{
    struct adjListNode* head;
};

struct Graph{
    int V;
    struct adjList* array;
};

struct adjListNode* newAdjListNode(int dest){
    struct adjListNode* newNode = (struct adjListNode*)malloc(sizeof(struct adjListNode));
    newNode->dest = dest;
    newNode->next = NULL;

    return newNode;
}


struct Graph* createGraph(int V){
    struct Graph* graph = (struct Graph*)malloc(sizeof(struct Graph));
    graph->V = V;

    graph->array = (struct adjList*)malloc(V * sizeof(struct adjList));

    int i;
    for(i = 0; i < V; i++){
        graph->array[i].head = NULL;
    }

    return graph;
}

void addEdge(struct Graph* graph, int src, int dest){
    struct adjListNode* newNode = newAdjListNode(dest);
    newNode->next = graph->array[src].head;
    graph->array[src].head = newNode;

    newNode = newAdjListNode(src);
    newNode->next = graph->array[dest].head;
    graph->array[dest].head = newNode;
}

void printGraph(struct Graph* graph)
{
    int v;
    for (v = 0; v < graph->V; ++v)
    {
        struct adjListNode* pCrawl = graph->array[v].head;
        printf("\n Adjacency list of vertex %d\n head ", v);
        while (pCrawl)
        {
            printf("-> %d", pCrawl->dest);
            pCrawl = pCrawl->next;
        }
        printf("\n");
    }
}

void DFS(struct Graph* graph, int v){
    if(visited[v] == 0){
        visited[v] = 1;
        printf("%d --> ", v);
        struct adjListNode *pCrawl = graph->array[v].head;
        while(pCrawl){
            if(visited[pCrawl->dest] == 0){
                DFS(graph, pCrawl->dest);
            }
            pCrawl = pCrawl->next;
        }
    }
}



int main(void) {
    int V = 5, I, src, dest, i, count = 0;
    scanf("%d %d", &V, &I);
    memset(visited, 0, SIZE);
    struct Graph* graph = createGraph(V);
    while(I--){
        scanf("%d %d", &src, &dest);
        addEdge(graph, src, dest);
    }

    for(i = 0; i < V; i++){
        if(visited[i] == 0){
            count++;
            DFS(graph, i);
            printf("\n");
        }
    }
    // print the adjacency list representation of the above graph
    printGraph(graph);
    printf("Countries :- %d", count);

    return 0;
}

输入 10 8 0 1 2 3 1 4 5 6 7 8 9 7 1 7 3 5

输出

SET :: 0 --> 1 --> 7 --> 9 --> 8 --> 4 --> 
SET :: 2 --> 3 --> 5 --> 6 --> 

 Adjacency list of vertex 0
 head -> 1

 Adjacency list of vertex 1
 head -> 7-> 4-> 0

 Adjacency list of vertex 2
 head -> 3

 Adjacency list of vertex 3
 head -> 5-> 2

 Adjacency list of vertex 4
 head -> 1

 Adjacency list of vertex 5
 head -> 3-> 6

 Adjacency list of vertex 6
 head -> 5

 Adjacency list of vertex 7
 head -> 1-> 9-> 8

 Adjacency list of vertex 8
 head -> 7

 Adjacency list of vertex 9
 head -> 7
Sets :- 2

Answer 1

您可以将数字表示为图形顶点。这对是边缘。现在您需要做的就是运行bfs来计算连接的部分。

Answer 2

Union-find algorithm对解决这个问题非常有效（你的例子不构建链，只有集合）

给定数字对可以创建所有链组

2 个答案: