解决负权边的算法(Bellman Ford )(有向图) (1)C ~

这个能够判断是否有负权边，但是不能计算有负圈的图，也就是说可以有负的权边，但是不能含有负权的环。

适用条件：

1.单源最短路径(从源点s到其它所有顶点v);

2.有向图&无向图(无向图可以看作(u,v),(v,u)同属于边集E的有向图);（如1 2  -3, 2 1 -3 两次输入 无向图）

3.边权可正可负(如有负权回路输出错误提示);

核心代码：

int bellman_ford(){for(int i = 1; i <= n-1; i++ ){for(int j = 1; j <= m; j++ ){if(dist[edge[j].v] > dist[edge[j].u] + edge[j].w){dist[edge[j].v] = dist[edge[j].u] + edge[j].w;pre[edge[j].v] = edge[j].u;}}}int flag = 0;//判断有无负值圈for(int j = 1; j <= m; j++ ){if(dist[edge[j].v] > dist[edge[j].u] + edge[j].w){flag = 1;}}return flag;}

完整实现：

#include<stdio.h>#include<stdlib.h>#define MAX 100#define INF 65535int n, m;int dist[MAX], pre[MAX];typedef struct Edge{int u;int v;int w;}Edge;Edge edge[MAX];int bellman_ford(){for(int i = 1; i <= n-1; i++ ){for(int j = 1; j <= m; j++ ){if(dist[edge[j].v] > dist[edge[j].u] + edge[j].w){dist[edge[j].v] = dist[edge[j].u] + edge[j].w;pre[edge[j].v] = edge[j].u;}}}int flag = 0;for(int j = 1; j <= m; j++ ){if(dist[edge[j].v] > dist[edge[j].u] + edge[j].w){flag = 1;}}return flag;}void print_path(int root){if(pre[root] != -1){print_path(pre[root]);printf("->");}printf("%d",root);}int main(){scanf("%d%d",&n, &m);for(int i = 1; i <= n; i++ ){dist[i] = INF;pre[i] = -1;}dist[1] = 0;//以从顶点1开始为例for(int j = 1; j <= m; j++ ){//输入边信息 scanf("%d%d%d",&edge[j].u, &edge[j].v, &edge[j].w);}printf("\n"); if(!bellman_ford()){for(int i = 1; i <= n; i++ ){if(dist[i] != INF){print_path(i);//打印路径printf("  distance = %d\n",dist[i]); printf("\n");}}}elseprintf("有负值圈.");}