问题:
/* * Copyright (c)2015,烟台大学计算机与控制工程学院 * All rights reserved. * 文件名称:项目5.cbp * 作 者:张芸嘉 * 完成日期:2015年12月4日 * 版 本 号:v1.0 * 问题描述:拓扑排序算法的验证 * 输入描述:无 * 程序输出:测试数据 */
代码:
#include "graph.h" void TopSort(ALGraph *G) { int i,j; int St[MAXV],top=-1; //栈St的指针为top ArcNode *p; for (i=0; i<G->n; i++) //入度置初值0 G->adjlist[i].count=0; for (i=0; i<G->n; i++) //求所有顶点的入度 { p=G->adjlist[i].firstarc; while (p!=NULL) { G->adjlist[p->adjvex].count++; p=p->nextarc; } } for (i=0; i<G->n; i++) if (G->adjlist[i].count==0) //入度为0的顶点进栈 { top++; St[top]=i; } while (top>-1) //栈不为空时循环 { i=St[top]; top--; //出栈 printf("%d ",i); //输出顶点 p=G->adjlist[i].firstarc; //找第一个相邻顶点 while (p!=NULL) { j=p->adjvex; G->adjlist[j].count--; if (G->adjlist[j].count==0)//入度为0的相邻顶点进栈 { top++; St[top]=j; } p=p->nextarc; //找下一个相邻顶点 } } } int main() { ALGraph *G; int A[10][10]= { {0,0,0,1,1,0,0,0,0,1}, {0,0,1,1,0,0,0,1,0,0}, {0,0,0,0,1,1,0,0,1,0}, {0,0,0,0,0,0,1,0,0,0}, {0,0,0,0,0,0,0,1,0,0}, {0,0,0,0,1,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,1,0,0,0}, {0,0,0,0,0,0,0,0,0,1}, {0,0,0,0,0,0,0,0,0,0} }; ArrayToList(A[0], 10, G); DispAdj(G); printf("\n"); printf("拓扑序列:"); TopSort(G); printf("\n"); return 0; }
测试用图:
运行结果:
知识点总结:
拓扑排序算法的验证。