4-1 Topological Sort (9分)

Write a program to find the topological order in a digraph.

Format of functions:

bool TopSort( LGraph Graph, Vertex TopOrder[] );

where LGraph is defined as the following:

typedef struct AdjVNode *PtrToAdjVNode; 
struct AdjVNode{
    Vertex AdjV;
    PtrToAdjVNode Next;
};

typedef struct Vnode{
    PtrToAdjVNode FirstEdge;
} AdjList[MaxVertexNum];

typedef struct GNode *PtrToGNode;
struct GNode{  
    int Nv;
    int Ne;
    AdjList G;
};
typedef PtrToGNode LGraph;

The topological order is supposed to be stored in TopOrder[] whereTopOrder[i] is the i-th vertex in the resulting sequence. The topological sort cannot be successful if there is a cycle in the graph -- in that caseTopSort must return false; otherwise return true.

Notice that the topological order might not be unique, but the judge's input guarantees the uniqueness of the result.

Sample program of judge:

#include 
#include 

typedef enum {false, true} bool;
#define MaxVertexNum 10  /* maximum number of vertices */
typedef int Vertex;      /* vertices are numbered from 0 to MaxVertexNum-1 */

typedef struct AdjVNode *PtrToAdjVNode; 
struct AdjVNode{
    Vertex AdjV;
    PtrToAdjVNode Next;
};

typedef struct Vnode{
    PtrToAdjVNode FirstEdge;
} AdjList[MaxVertexNum];

typedef struct GNode *PtrToGNode;
struct GNode{  
    int Nv;
    int Ne;
    AdjList G;
};
typedef PtrToGNode LGraph;

LGraph ReadG(); /* details omitted */

bool TopSort( LGraph Graph, Vertex TopOrder[] );

int main()
{
    int i;
    Vertex TopOrder[MaxVertexNum];
    LGraph G = ReadG();

    if ( TopSort(G, TopOrder)==true )
        for ( i=0; iNv; i++ )
            printf("%d ", TopOrder[i]);
    else
        printf("ERROR");
    printf("\n");

    return 0;
}

/* Your function will be put here */

Sample Input 1 (for the graph shown in the figure):

5 7
1 0
4 3
2 1
2 0
3 2
4 1
4 2

Sample Output 1:

4 3 2 1 0

Sample Input 2 (for the graph shown in the figure):

5 8
0 3
1 0
4 3
2 1
2 0
3 2
4 1
4 2

Sample Output 2:

ERROR

bool TopSort(LGraph Graph, Vertex TopOrder[]) {
	int i=0,count=0,j;
	int degree[MaxVertexNum] = {0};
	int deque[MaxVertexNum];
	int front=0, rear=0;
	PtrToAdjVNode temp;
	for(i=0;iNv;i++){              //找到节点的度
		temp = Graph->G[i].FirstEdge;
		while (temp != NULL) {
			degree[temp->AdjV]++;
			temp = temp->Next;			
		}
	}
	for (j = 0; j < Graph->Nv; j++) {
		if (degree[j] == 0) {
			deque[rear++] = j;
			break;
		}
	}
	while (rear != front) {        //用队列不会超时
		TopOrder[count++] = deque[front];
		temp = Graph->G[deque[front]].FirstEdge;
		while (temp != NULL) {
			degree[temp->AdjV]--;
			if (degree[temp->AdjV] == 0)
				deque[rear++] = temp->AdjV;
			temp = temp->Next;
		}
		front++;
	}
	/*for (i = 0; i < Graph->Nv; i++) {//这是原来写的暴力解法，某个点会超时，看来还是需要多加锻炼思维
		vno = Graph->Nv;
		for (j = 0; j < Graph->Nv; j++) {
			if (a[j] == 0) {
				vno = j;
				break;
			}
		}
		if (vno >= Graph->Nv)
			return false;
		else {
			TopOrder[i] = vno;
			a[vno] = MaxVertexNum+1;
			temp = Graph->G[vno].FirstEdge;
			while (temp != NULL) {
				a[temp->AdjV]--;
				temp = temp->Next;

			}
		}
	}*/
	if(count==Graph->Nv)
		return true;
	return false;
}

感想：

1.注意图示意思，FirstEdge指的是第一条边而不是链表的起始点

2.注意算法使用，其实用会更好