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 <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 */
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;i<Graph->Nv;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<deque>会更好