算法导论 所有节点对的最短路径问题 矩阵法

#include 
#include 
#include 

//图节点  
typedef struct VertexNode  
{  
    char name;  
    VertexNode *p; 
}Vertex,*pVertex;  
  
//图  
typedef struct   
{  
    int vn;  
    int **E;  
    pVertex *V;  
      
}Graph,*pGraph;  
//根据算法导论 图25-1 初始化图  
pGraph initGraph()  
{  
    pGraph g=(pGraph)malloc(sizeof(Graph));  
    g->vn=5;  
    pVertex v1=(pVertex)malloc(sizeof(Vertex));  
    v1->name='1';    
    v1->p=NULL;  
    pVertex v2=(pVertex)malloc(sizeof(Vertex));  
    v2->name='2';    
    v2->p=NULL;  
    pVertex v3=(pVertex)malloc(sizeof(Vertex));  
    v3->name='3';   
    v3->p=NULL;  
    pVertex v4=(pVertex)malloc(sizeof(Vertex));  
    v4->name='4';    
    v4->p=NULL;  
    pVertex v5=(pVertex)malloc(sizeof(Vertex));  
    v5->name='5';   
    v5->p=NULL;  
  
    g->V=(pVertex*)malloc(g->vn*sizeof(pVertex));  
    g->V[0]=v1;  
    g->V[1]=v2;  
    g->V[2]=v3;  
    g->V[3]=v4;  
    g->V[4]=v5;  
  
    g->E = (int**)malloc(g->vn*sizeof(int*));  
    for(int i=0;ivn;i++)  
    {  
        g->E[i]=(int*)malloc(g->vn*sizeof(int));  
    }  
    for(int i=0;ivn;i++)  
    {  
        for(int j=0;jvn;j++)  
        { 
			if(i==j)
				g->E[i][j]=0;
			else
				g->E[i][j]=INT_MAX;  
        }  
    }  
    g->E[0][1]=3;  
    g->E[0][2]=8;  
	g->E[0][4]=-4;
    g->E[1][3]=1;  
    g->E[1][4]=7;  
    g->E[2][1]=4;  
    g->E[3][2]=-5;
	g->E[3][0]=2;
    g->E[4][3]=6;  
    return g;  
}  

int ** ExtendShortestPaths(int **L,int **W,int n)
{
	int **P=(int**)malloc(n*sizeof(int*));
	for(int i=0;isum)
					P[i][j]=sum;
			}
		}
	}
	return P;
}

int ** AllPathShortestPaths(int **W,int n)
{
	int **L=(int**)malloc(n*sizeof(int*)),**temp;
	for(int i=0;iE,g->vn);
	printRec(L,g->vn);
	getchar();
}