C

Question

  

说明
  用C编写程序以使用邻接矩阵实现图   技术。向用户询问节点和边的数量。然后要求   节点标签，边缘标签。最后询问源节点和目标节点   每个。根据输入，您需要构造图形并   执行以下操作。

     

  
1. 如果用户输入节点标签，则应该   输出该节点的进出度。
  
2. 如果用户输入边缘   标签，它应该输出其源节点和目标节点。
  
3. 如果用户   输入一个节点标签，它应该输出该节点的所有入射边   节点。
  

#include <stdio.h>

#define m 50

int ad_mat[m][m];
int n;

int graph(void){

    int i, j, source, destination, nd, edge, in_deg, out_deg;

    printf("Enter number of nodes:");
    scanf("%d", &n);

    printf("\n Enter the lables for nodes:");
    for (i = 1; i <= n; i++) {
        scanf("%d", &nd);
    }

    printf("\n Enter the number of edges:");
    scanf("%d", &edge);

    int edgear[edge];
    printf("\n Labling edges");
    for (i = 1; i <= edge; i++) {
        scanf("%d", &edgear[i]);
        printf("\n");
    }

    //labling the edges
    for (i = 1; i <= edge; i++) {
        printf("\nEnter the source and destination for node %d:", edgear[i]);
        scanf("%d %d", &source, &destination);
        if ((source) == 0 && (destination) == 0) {
            break;
        }

        if (source > n || destination > n || source <= 0 || destination <= 0) {
            printf("\nInvalid edge!");
            i--;
        }
        else{
            ad_mat[source][destination] = 1;
        }

    }
    //print adjacency matric
    printf("\n adjacency matrix for directed graph\n");
    for (i =1; i <= n; i++) {
        for (j = 1; j <= n; j++) {
            printf("%4d", ad_mat[i][j]);
        }
        printf("\n");
    }

    //For in-degree and out-degree
    for (i = 1; i <= n; i++) {
        in_deg = out_deg =0;
        for (j =1; j <= n; j++) {
            if (ad_mat[j][i] == 1) {
                in_deg++;
            }
        }
        for (j = 1; j <= n; j++) {
            if (ad_mat[i][j] == 1) {
                out_deg++;
            }
        }
    }




    return 0;
}

/*int display(){
    int i,j;
    for (i =1; i <= n; i++) {
        for (j = 1; j <= n; j++) {
            printf("%4d", ad_mat[i][j]);
        }
        printf("\n");
    }
    return 0;
}*/
int main(int argc, const char * argv[]) {

    graph();
   // display();


    return 0;
}