Floyd算法C语言实现

<pre name="code" class="objc">#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define NUM 4

typedef struct MGraph		/* 邻接表存储结构 */
{
int edges[NUM][NUM];
int n,e;
} MGraph;

MGraph *build_mgraph();
void Floyd(MGraph *mgraph);
void Ppath(int path[][NUM], int i, int j);
void Dispath(int A[][NUM], int path[][NUM], int n);

int main(void)
{
MGraph *mgraph;

printf("\n*************************************************************\n");
printf("该图的矩阵表示为:\n");
mgraph=build_mgraph();
printf("\n*************************************************************\n");
printf("各顶点间最短路径为:\n");
Floyd(mgraph);
printf("\n*************************************************************\n");

return 0;
}

MGraph *build_mgraph()
{
int i,j;
int num_e=0;
MGraph *mgraph=(MGraph *)malloc(sizeof(MGraph));
int matrix[NUM][NUM]={{0,5,INT_MAX,7},
{INT_MAX,0,4,2},
{3,3,0,2},
{INT_MAX,INT_MAX,1,0}};
for(i=0;i<NUM;i++)
{
for(j=0;j<NUM;j++)
{
mgraph->edges[i][j]=matrix[i][j];
if(matrix[i][j]!=0 && matrix[i][j]!=INT_MAX)
num_e++;
}
}
mgraph->n=NUM;
mgraph->e=num_e;

printf("node=%d;edges=%d\n",mgraph->n,mgraph->e);
for(i=0;i<NUM;i++)
{
for(j=0;j<NUM;j++)
{
if(mgraph->edges[i][j]!=INT_MAX)
printf("%3d",mgraph->edges[i][j]);
else
printf("%3c",'&');
}
printf("\n");
}

return mgraph;
}

void Floyd(MGraph *mgraph)
{
int A[NUM][NUM],path[NUM][NUM];
int i,j,k;

for(i=0;i<mgraph->n;i++)
{
for(j=0;j<mgraph->n;j++)
{
A[i][j]=mgraph->edges[i][j];
path[i][j]=-1;
}
}

for(k=0;k<mgraph->n;k++)
{
for(i=0;i<mgraph->n;i++)
{
for(j=0;j<mgraph->n;j++)
{
if(A[i][k]!=INT_MAX && A[k][j]!=INT_MAX && A[i][j]>A[i][k]+A[k][j])
{
A[i][j]=A[i][k]+A[k][j];
path[i][j]=k;
}
}
}
}

Dispath(A,path,mgraph->n);
}

void Ppath(int path[][NUM], int i, int j)
{
int k;

k=path[i][j];
if(k==-1)
return;
Ppath(path,i,k);
printf("%d,",k);
Ppath(path,k,j);
}
void Dispath(int A[][NUM], int path[][NUM], int n)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(A[i][j]==INT_MAX)
printf("%d-%d have no path",i,j);
printf("%d-%d-%d: ",i,j,A[i][j]);
printf("%d,",i);
Ppath(path,i,j);
printf("%d\n",j);
}
}
}