Floyd算法解决多源最短路径问题

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define max 100
#define INF 999
int graph[max][max];
int vertex_num;
int edge_num;
int d[max][max];
int p[max][max];

void Floyd(){
    int i,j,k;
    for(i=0;i<vertex_num;i++){
        for(j=0;j<vertex_num;j++){
            d[i][j]=graph[i][j];
            p[i][j]=-1;
        }
    }
    for(k=0;k<vertex_num;k++){
        for(i=0;i<vertex_num;i++){
            for(j=0;j<vertex_num;j++){
                if(d[i][j]>d[i][k]+d[k][j]){
                    d[i][j]=d[i][k]+d[k][j];
                    p[i][j]=k;
                }
            }
        }
    }    
    
}
void find_path(int i,int j){
    int k;
    k=p[i][j];
    if(k==-1)return;
    find_path(i,k);
    printf("%d ",k);
    find_path(k,j);
}
void show_path(){
    int i,j;
    
    printf("Output:\n");
    for(i=0;i<vertex_num;i++){
        for(j=0;j<vertex_num;j++){
            if(d[i][j]==INF){
                if(i!=j)printf("No path from %d to %d\n",i,j);
            }else{
                printf("Path from %d to %d: ",i,j);
                printf("%d ",i);
                find_path(i,j);
                printf(" %d",j);
                printf(" distance:%-5d\n",d[i][j]);
            }
        }
        printf("\n");
    }
}

int main(){
    int i,j;
    FILE *fin  = fopen ("dij.in", "r");
    FILE *fout = fopen ("dij.out", "w");
    
    char buf[10];
    fgets(buf,10,fin);
    edge_num=atoi(buf);
    
    printf("edge_num:%d\n",edge_num);
    fgets(buf,10,fin);
    vertex_num=atoi(buf);

    printf("vertex_num:%d\n",vertex_num);
    
    for(i=0;i<edge_num;i++){
        int start,end,weight;//start point,end point and the weight of edge
        fgets(buf,10,fin);
        sscanf(buf,"%d %d %d",&start,&end,&weight);
        
        printf("start:%d end:%d weight:%d\n",start,end,weight);
        graph[start][end]=weight;//init the graph matrix 
        
    }

    Floyd();
    show_path();
    return 0;
}