floyd算法的C语言实现

int *s ;//定义具有最短路径的节点子集s
s = (int *)malloc(sizeof(int) * n);
//初始化最小路径代价和前一跳节点值
for (i = 1; i <= n; i++)
{
dist[i] = cost[v][i];
s[i] = 0;
if (dist[i] == maxint)
MGraph G;
int path[MAX_VERTEX_NUM][MAX_VERTEX_NUM];//v到各顶点的最短路径向量
int D[MAX_VERTEX_NUM][MAX_VERTEX_NUM];//v到各顶点最短路径长度向量
//初始化
AdjType a[MAX_VERTEX_NUM]Байду номын сангаасMAX_VERTEX_NUM]={
printf("(V%d",i);//输出Vi的序号i
while(k!=j){//k不等于路径终点j时
printf(",V%d",k);//输出k
k=path[k][j];//求路径上下一顶点序号
}
printf(",V%d)\n",j);//输出路径终点序号
}
printf("\n");
}
}
system("pause");
{
prev[i] = 0;
}
else
{
prev[i] = v;
}
}
dist[v] = 0;
s[v] = 1;//源节点作为最初的s子集
for (i = 1; i < n; i++)
{
int temp = maxint;
int u = v;
//加入具有最小代价的邻居节点到s子集
for (j = 1; j <= n; j++)
}
}
dist = (int *)malloc(sizeof(int)*n);
prev = (int *)malloc(sizeof(int)*n);
printf("please input the source node: ");
scanf("%d",&v);
//调用dijkstra算法
Dijkstra(n, v, dist, prev, cost);
//Floyd算法
//求网G（用邻接矩阵表示）中任意两点间最短路径
//D[][]是最短路径长度矩阵，path[][]最短路径标志矩阵
void Floyd(MGraph * G,int path[][MAX_VERTEX_NUM],int D[][MAX_VERTEX_NUM],int n){
int i,j,k;
{
int newdist = dist[u] + cost[u][j];
if (newdist < dist[j])
{
dist[j] = newdist;
prev[j] = u;
}
}
}
}
}
//主函数，主要做输入输出工作
void main()
{
int i,j,t;
int n,v,u;
int **cost;//代价矩阵
for(i=0;i<n;i++){//初始化
for(j=0;j<n;j++){
if(G->A[i][j]<MAX_INT){
path[i][j]=j;
}else{
path[i][j]=-1;
}
D[i][j]=G->A[i][j];
}
}
for(k=0;k<n;k++){//进行n次试探
for(i=0;i<n;i++){
};
for(i=0;i<n;i++){
for(j=0;j<n;j++){
G.A[i][j]=a[i][j];
}
}
Floyd(&G,path,D,6);
for(i=0;i<n;i++){//输出每对顶点间最短路径长度及最短路径
for(j=0;j<n;j++){
printf("V%d到V%d的最短长度:",i,j);
{
if ((!s[j]) && (dist[j] < temp))
{
u = j;
temp = dist[j];
}
}
s[u] = 1;
//计算加入新的节点后，更新路径使得其产生代价最短
for (j = 1; j <= n; j++)
{
if ((!s[j]) && (cost[u][j] < maxint))
printf("%d\t",D[i][j]);//输出Vi到Vj的最短路径长度
k=path[i][j];//取路径上Vi的后续Vk
if(k==-1){
printf("There is no path between V%d and V%d\n",i,j);//路径不存在
}else{
printf("最短路径为:");
for(j=0;j<n;j++){
if(D[i][j]>D[i][k]+D[k][j])
{
D[i][j]=D[i][k]+D[k][j];//取小者
path[i][j]=path[i][k];//改Vi的后继
}
}
}
}
}
int main(){
int i,j,k,v=0,n=6;//v为起点，n为顶点个数
{0,12,18,MAX_INT,17,MAX_INT},
{12,0,10,3,MAX_INT,5},
{18,10,0,MAX_INT,21,11},
{MAX_INT,3,MAX_INT,0,MAX_INT,8},
{17,MAX_INT,21,MAX_INT,0,16},
{MAX_INT,5,11,8,16,0}
for (i = 1; i <= n; i++)
{
cost[i]=(int *)malloc(sizeof(int)*(n+1));
}
//输入代价矩阵
for (j = 1; j <= n; j++)
{
for (t = 1; t <= n; t++)
{
scanf("%d",&cost[j][t]);
}
int *dist;//最短路径代价
int *prev;//前一跳节点空间
printf("please input the node number: ");
scanf("%d",&n);
printf("please input the cost status:\n");
cost=(int **)malloc(sizeof(int)*(n+1));
return 0;
}
//Dijkstra算法实现函数
＃i nclude<stdio.h>
＃i nclude <stdlib.h>
void Dijkstra(int n,int v,int dist[],int prev[],int **cost)
{
int i;
int j;
int maxint = 65535;//定义一个最大的数值，作为不相连的两个节点的代价权值