我想从给定的源和目标打印所有可能的路径。但是在我的BFS代码中,它仅显示两条路径,而不显示多条路径。对于给定图,其中n = 4,边= 6,给定,
1-2
1-3
1-5
5-3
5-4
3-4
3-2
它应该已经打印了3条路径:
1-5-4
1-3-4
1-5-3-4
但是它只显示了这两个路径
1-3-4
1-5-4
这是我的示例代码,用于查找到目标路径的src
#include <stdio.h>
int queue1[100], state[100], parent[100];
int front = 0, rear = -1, maxSize = 100;
int count = 0;
int initial = 1, waiting = 2, visited = 3;
int n, e;
int adj[100][100];
bool isEmpty()
{
return count == 0;
}
bool isFull()
{
return count == maxSize;
}
void enqueue(int val)
{
if (!isFull())
{
if (rear == maxSize - 1)
{
rear = -1;
}
rear++;
queue1[rear] = val;
count++;
}
}
int dequeue()
{
int val = queue1[front];
front++;
if (front == maxSize)
{
front = 0;
}
count--;
return val;
}
void BFS_Traversal(int src, int des)
{
int done = 0;
enqueue(src);
state[src] = waiting;
parent[src] = -1;
printf("path ");
while (!isEmpty() && done == 0)
{
src = dequeue();
// printf("%d ",src);
state[src] = visited;
for (int i = 1; i <= n; i++)
{
if (adj[src][i] == 1 && state[i] == initial)
{
enqueue(i);
state[i] = waiting;
parent[i] = src;
if (i == des)
{
state[i] = initial;
int k = des;
do
{
printf("%d ", k);
k = parent[k];
} while (k != -1);
printf("\n");
}
}
}
}
}
int main()
{
int src, start, end, des;
scanf("%d%d", &n, &e);
for (int i = 1; i <= e; i++)
{
scanf("%d%d", &start, &end);
adj[start][end] = 1;
}
for (int i = 1; i <= n; i++)
{
state[i] = initial;
}
for (int k = 1; k <= n; k++)
{
parent[k] = -1;
}
scanf("%d%d", &src, &des);
BFS_Traversal(src, des);
}
因此,您可以看到未显示1-5-3-4路径,因为它们已被访问。我应该如何修改此代码以打印所有可能的路径?