深度优先遍历找出一个无向图中的环

进行深度优先遍历的时候,当考察的点的下一个邻接点是已经被遍历的点,并且不是自己之前的父亲节点的时候,我们就找到了一条逆向边,因此可以判断该无向图中存在环路。

visited数组记录了节点的访问状态,visited[i] = 0表示节点i尚未被访问过; visited[i] = 1表示节点i被访问了,但是尚未被检测完毕; visited[i] = 2表示节点i已经被检测完毕了,对于检测完毕了的节点,其所有的临边都已经被考虑过了,如果存在逆向边的话已经被检测了,从而避免了重复地进行检测。 father[i]表示到达节点i经过的前驱节点,通过反向索引father数组,我们可以找出这个环。

#include <stdio.h>
#include <stdlib.h>

int * newIntRaw(int n)
{
    return (int *)malloc(sizeof(int) * n);
}
int * newInt(int n, int init)
{
    int *p = newIntRaw(n);
    int i;
    for (i = 0; i < n; ++i)
        p[i] = init;
    return p;
}
int ** newMap(int n, int m, int init)
{
    int **res = (int **)malloc(sizeof(int *) * n);
    int i;
    for (i = 0; i < n; ++i)
        res[i] = newInt(m, init);
    return res;
}

typedef struct
{
    int e;
    int n;
    int **map;
} Graph;

Graph * newGraph()
{
    int n, e;
    int i;
    int from, to;
    Graph *g = (Graph *)malloc(sizeof(Graph));

    scanf("%d %d", &n, &e);
    g->n = n;
    g->e = e;
    g->map = newMap(n, n, 0);
    for (i = 0; i < e; ++i) {
        scanf("%d %d", &from, &to);
        g->map[from][to] = 1;
        g->map[to][from] = 1;
    }
    return g;
}

void dispGraph(Graph *g)
{
    int i, j;
    for (i = 0; i < g->n; ++i) {
        printf("%d: ", i);
        for (j = 0; j < g->n; ++j) {
            if (g->map[i][j] == 1)
                printf("%d ", j);
        }
        printf("\n");
    }
}

// ---------------------------solve-----------------------
int *visited;
int *father;

void dfs(Graph *g, int s)
{
    int i;
    visited[s] = 1;
    for (i = 0; i < g->n; ++i) {
        if (g->map[s][i] == 1) {
            if (visited[i] == 0) {
                father[i] = s;
                dfs(g, i);
            }
            else if (visited[i] == 1 && i != father[s]) {
                int tmp = s;
                printf("find a ring: %d --> ", i);
                while (tmp != i) {
                    printf("%d --> ", tmp);
                    tmp = father[tmp];
                }
                printf("%d\n", tmp);
            }
        }
    }
    visited[s] = 2; // 避免重复计算环
}

void findRing(Graph *g)
{
// dispGraph(g);
    int i;
    visited = newInt(g->n, 0);
    father = newInt(g->n, -1);
    for (i = 0; i < g->n; ++i)
        if (visited[i] == 0)
            dfs(g, i);
}

int main()
{
    Graph *g = newGraph();
    findRing(g);
    return 0;
}

/* 5 6 0 1 0 4 4 3 3 1 1 2 3 2 */
    原文作者:数据结构之图
    原文地址: https://blog.csdn.net/dfq12345/article/details/78004876
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