Java C 实现Bellman-ford算法

package com.cn.tree;public class MyException extends Exception {private static final long serialVersionUID = 1L;public MyException(String str) {super(str);}public MyException() {}}

package com.cn.graph;import com.cn.tree.MyException;/** * 很明显时间复杂度为O(VE),因为每一次需要对边进行松弛，所以我们采用保存边的方式来存储图的的信息。 * p保存的是前驱节点，d保存的是源点到每一个点的最短距离。我们最后又做了一次判断，如果还有可以松弛 * 的边，那么可以保证的是图中有负权的环存在，这样的话就没有最短路径只说了，可以无限小。 *  * @author daijope * */public class BellmanFord {private static final int m = 10000;public static void main(String[] args) throws MyException {Adge a1 = new Adge(0, 1, -1);Adge a2 = new Adge(0, 2, 4);Adge a3 = new Adge(1, 2, 3); Adge a4 = new Adge(3, 1, 1);Adge a5 = new Adge(1, 3, 2);Adge a6 = new Adge(3, 2, 5);Adge a7 = new Adge(1, 4, 2);Adge a8 = new Adge(4, 3, -3);Adge[] as = new Adge[]{a1, a2, a3, a4, a5, a6, a7, a8};int[] d = new int[5];int[] p = new int[5];d[0] = 0;p[0] = 0;for (int i = 1; i < d.length; i++) {d[i] = m;p[i] = -1;}bellman_Ford(as, d, p);for (int i = 0; i < d.length; i++) {System.out.println(d[i]);}}private static void bellman_Ford(Adge[] as, int[] d, int[] p) throws MyException {for(int i = 1; i < d.length; i++) {for (int j = 0; j < as.length; j++) {if (d[as[j].getV()] > d[as[j].getU()] + as[j].getW()) {d[as[j].getV()] = d[as[j].getU()] + as[j].getW();p[as[j].getV()] = as[j].getU();}}}for (int j = 0; j < as.length; j++) {if (d[as[j].getV()] > d[as[j].getU()] + as[j].getW()) {throw new MyException("有负环");}}}}class Adge {private int u;private int v;private int w;public Adge(int u, int v, int w) {this.u = u;this.v = v;this.w = w;}public int getU() {return u;}public void setU(int u) {this.u = u;}public int getV() {return v;}public void setV(int v) {this.v = v;}public int getW() {return w;}public void setW(int w) {this.w = w;}}

c语言实现：

#include<stdio.h>#define MAXVALUE 10000typedef struct node{     int u;     int v;     int w;};bool BELLMAN_FORD(node G[]){         int dis[5],pre[5],p[5];    int i,j,k;    for(i=0;i<=4;i++)    {        dis[i]=MAXVALUE;        pre[i]=-1;        p[i]=1;    }    dis[G[0].u]=0;  for(j=1;j<=4;j++)    {        for(i=0;i<=7;i++)        {           if(dis[G[i].v]>dis[G[i].u]+G[i].w)           {               dis[G[i].v]=dis[G[i].u]+G[i].w;               pre[G[i].v]=G[i].u;           }             }                   }     for(i=1;i<=7;i++)     {          if(dis[G[i].v]>dis[G[i].u]+G[i].w)                            return false;          }     for(i=0;i<=7;i++)    {               if(p[G[i].v]==1)           {            printf("%d   %d\n",G[i].v,dis[G[i].v]);            k=G[i].v;            p[G[i].v]=0;            while(pre[k]!=-1)            {                printf("%d   \n",pre[k]);                k=pre[k];            }          }    }    return true;}int main(){    node G[8];    G[0].u=0;    G[0].v=1;    G[0].w=-1;      G[1].u=0;    G[1].v=2;    G[1].w=3;    G[2].u=1;    G[2].v=2;    G[2].w=3;    G[3].u=1;    G[3].v=3;    G[3].w=2;    G[4].u=1;    G[4].v=4;    G[4].w=2;    G[5].u=3;    G[5].v=1;    G[5].w=1;    G[6].u=3;    G[6].v=2;    G[6].w=5;    G[7].u=4;    G[7].v=3;    G[7].w=-3;      BELLMAN_FORD(G);    getchar();        }//O(VE)