Write a program to find the topological order in a digraph.
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 <stdio.h>
#include <stdlib.h>
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; i<G->Nv; i++ )
printf("%d ", TopOrder[i]);
else
printf("ERROR");
printf("\n");
return 0;
}
/* Your function will be put here */
5 7
1 0
4 3
2 1
2 0
3 2
4 1
4 2
Sample Output 1:
4 3 2 1 0
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 begin=0,end=0; int i,k=0,c=0; int q[MaxVertexNum]; int indegree[MaxVertexNum]; PtrToAdjVNode a; for(i=0;i<MaxVertexNum;i++) //在pta里用{0}归零数组会报错,只能如此归零了 indegree[i]=0; for(i=0;i<Graph->Nv;i++) { //遍历节点初始化入度数组 a=Graph->G[i].FirstEdge; //G[i],就是每个结点的邻接表,a指向i顶点指向的下一个顶点 while(a!=NULL){ indegree[a->AdjV]++; //如果i顶点的下一个顶点存在,则此顶点入度减一 a=a->Next; } if(indegree[i]==0) //寻找入度为零的结点 q[end++]=i; } while(begin!=end){ //只有用队列才不超时 TopOrder[k++]=q[begin++]; a=Graph->G[q[begin-1]].FirstEdge; while(a!=NULL){ //更新a周围的顶点的入度并且找到为零的入队 indegree[a->AdjV]--; if(indegree[a->AdjV]==0) q[end++]=a->AdjV; a=a->Next; } } if(Graph->Nv!=k) //当队列为空的时候还有没入队的顶点 return false; return true; }
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