图论算法：拓扑排序（Topological Sort）

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拓扑在计算机科学中的意思是能够把对象进行安排，使得它们可以被边连接。

从一个偏序（自反的、反对称的和传递的关系）构造一个相容的全序（集合中每对元素都可比，即每对元素都满足所定义的偏序关系R）叫做拓扑排序，这里的R可以定义为：元素a,b是需要执行的任务，b开始当且仅当a结束。

下面是伪代码：

Procedure topological sort((S, <<): finite partially ordered set)

k:=1

while S <> ∅

ak:=the minimal of S

S:=S-{ak}

k:=k+1

return a1,a2,a3,...,an{a1,a2,a3,...,an is a topological order}

这样做的道理是极小元一定存在。

画出Hasse图以后很容易找到拓扑排序。

在编程的时候需要找到入度为0的点（优先级最低），没有这种点，这个图一定有环。

4-13 Topological Sort (25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[] 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 <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 71 04 32 12 03 24 14 2

Sample Output 1:

4 3 2 1 0

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

5 80 31 04 32 12 03 24 14 2

Sample Output 2:

ERROR

因为学校诚信守则的关系不让贴代码，记录一下解题思路供自己复习：

1）需要用一个数组记录每个点的入度

2）遍历这个数组，度为0的加到栈里

3）出栈，把这个顶点放到记录顺序的数组，在邻接表中找到这个顶点，把与它邻接的顶点的度减1，如果减了以后度为0，入栈

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