邻接矩阵 有向图 判断是否有环 是否连通 DFS C实现~

1 DFS 搜索 修改

            有向图有环需要注意 ：按照路径方向能组成回路才叫有环

              例如 三边 A->B, A->C, B->C则 构成的不是环

                          A->B,B->C,C-A 组成的才是环

// 因为是有向图两个顶点也可以成环 void dfs(Graph g, int i){int j;color[i] = gray; //灰色 表示正在访问printf("%c ", g.vex[i]);for (j = 0; j < g.vex_num; j++) {if (i != j && g.edge[i][j] != INFINITY) {//两顶点有边相连if (color[j] == white) {dfs(g, j);//如果该节点未访问 继续访问其临近节点}else if(color[j] == gray){loop_num++;is_dag = false; //有环}}}color[i] = black;}void dfs_trvsal(Graph g){int i;for (i = 0; i < g.vex_num; i++) {color[i] = white;loop_num = 0;}link_component = 0;for (i = 0; i < g.vex_num; i++) {if (color[i] == white) {link_component++;dfs(g, i);}}}

完整实现 ：

#include<stdio.h>#include<stdlib.h>#include<ctype.h>#define MAXVEX 100#define INFINITY 65535#define TRUE 1#define FALSE 0#define gray -1#define white 0#define black 1int loop_num; //环 数量int color[MAXVEX];  //-1 表示灰色  0表示白色 1 表示黑色bool is_dag = true;//是否是有向无环图int link_component; // 连通分量 数量typedef int EdgeType;typedef char VertexType;typedef struct Graph {VertexType vex[MAXVEX];EdgeType edge[MAXVEX][MAXVEX];int vex_num, edge_num;}Graph;void init_graph(Graph *g){int i, j;for (i = 0; i < g->vex_num; i++) {for (j = 0; j < g->vex_num; j++) {if (i == j) {g->edge[i][j] = 0;}elseg->edge[i][j] = INFINITY;}}}char read_char(){char ch;do {ch = getchar();} while (!isalpha(ch));return ch;}int get_pos(Graph g, char ch){int i;for (i = 0; i < g.vex_num; i++) {if (g.vex[i] == ch)return i;}return -1;}void create_graph(Graph *g){int i, k;//printf("请输入顶点数与边数:\n");scanf("%d%d", &g->vex_num, &g->edge_num);init_graph(g);// 初始化  //printf("请输入顶点信息:\n");for (i = 0; i < g->vex_num; i++) {//scanf("%c", g->vex[i]);g->vex[i] = read_char();}//printf("请输入边的信息:\n");char c1, c2;int p1, p2,w;for (k = 0; k < g->edge_num; k++) {c1 = read_char();c2 = read_char();scanf("%d", &w);p1 = get_pos(*g, c1);p2 = get_pos(*g, c2);g->edge[p1][p2] = w;//有向边的权重}}void print_graph(Graph g){int i, j;for (i = 0; i < g.vex_num; i++) {for (j = 0; j < g.vex_num; j++) {if (g.edge[i][j] == INFINITY)printf("%5c", '*');else {printf("%5d", g.edge[i][j]);}}printf("\n");}}// 因为是有向图两个顶点也可以成环 void dfs(Graph g, int i){int j;color[i] = gray; //灰色 表示正在访问printf("%c ", g.vex[i]);for (j = 0; j < g.vex_num; j++) {if (i != j && g.edge[i][j] != INFINITY) {//两顶点有边相连if (color[j] == white) {dfs(g, j);//如果该节点未访问 继续访问其临近节点}else if(color[j] == gray){loop_num++;is_dag = false; //有环}}}color[i] = black;}void dfs_trvsal(Graph g){int i;for (i = 0; i < g.vex_num; i++) {color[i] = white;loop_num = 0;}link_component = 0;for (i = 0; i < g.vex_num; i++) {if (color[i] == white) {link_component++;dfs(g, i);}}}int main(){Graph g;create_graph(&g);//print_graph(g);printf("\n\nDFS:\t");dfs_trvsal(g);printf("\n\n");if (is_dag) {printf("该有向图无环.\n");}else {printf("该有向图有%d个环.\n", loop_num);}if (link_component == 1) {printf("该有向图连通.");}else {printf("该有向图不连通.");}getchar();getchar();return 0;}