6-16 Topological Sort（25 point(s)）

6-16 Topological Sort（25 point(s)）

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[] where TopOrder[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 case TopSort 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

拓扑排序问题，具体讲解可以看点击打开链接 ，感觉讲的很清楚
代码：

bool TopSort( LGraph Graph, Vertex TopOrder[] ){
	int indegree[MaxVertexNum];//用于保存每个点的入度 
	int q[MaxVertexNum];//队列，用于优化时间复杂度，每次找零度的点只需要从上次零度点相邻的里面找
	int i;
	int num = 0;//记录拓扑排序数组下标 
	int head=0,tail=0;
	PtrToAdjVNode t;
	for(i = 0; i < Graph->Nv; i++){
		indegree[i] = 0;
	}
	for(i = 0; i < Graph->Nv; i++){//遍历每一个点 
		t = Graph->G[i].FirstEdge; 
		while(t){//遍历每一个点后面所连的点 
			indegree[t->AdjV]++;
			t = t->Next;
		}
	}//对入度进行初始化 
	for(i = 0; i < Graph->Nv; i++){
		if(indegree[i]==0){
			q[tail++] = i;//先找到入度为零的点，加入队列，然后对队列进行操作即可 
		}
	}
	if(head==tail)return false;//如果找了一圈没有入度为零的点，肯定不对，返回false
	//队列操作
	while(headG[q[head]].FirstEdge;//取出队首的元素
		while(t){//对队首元素相邻的点进行遍历，操作 
			if(indegree[t->AdjV]<0)return false;//如果这个点入度为负数，说明之前肯定已经放在拓扑排序数组里了，再出现说明有环 
			indegree[t->AdjV]--;//相邻元素入度减一 
			if(indegree[t->AdjV]==0){
				q[tail++] = t->AdjV;//减后为零入队 
			}
			t = t->Next;
		} 
		indegree[q[head]] = -1;//表示这个点不能再出现了 
		TopOrder[num++] = q[head++]; 
	}
	if(num!=Graph->Nv)return false;
	else return true; 
}