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2.cbp * 作 者:张芸嘉 * 完成日期:2015年11月27日 * 版 本 号:v1.0 * 问题描述:假设图G采用邻接表存储,分别设计实现以下要求的算法: * 输入描述:无 * 程序输出:测试数据 */
(图1)
代码:
(1)输出出图G中每个顶点的出度
#include "graph.h" //返回图G中编号为v的顶点的出度 int OutDegree(ALGraph *G,int v) { ArcNode *p; int n=0; p=G->adjlist[v].firstarc; while (p!=NULL) { n++; p=p->nextarc; } return n; } //输出图G中每个顶点的出度 void OutDs(ALGraph *G) { int i; for (i=0; i<G->n; i++) printf(" 顶点%d:%d\n",i,OutDegree(G,i)); } int main() { ALGraph *G; int A[7][7]= { {0,1,1,1,0,0,0}, {0,0,0,0,1,0,0}, {0,0,0,0,1,1,0}, {0,0,0,0,0,0,1}, {0,0,0,0,0,0,0}, {0,0,0,1,1,0,1}, {0,1,0,0,0,0,0} }; ArrayToList(A[0], 7, G); printf("各顶点出度:\n"); OutDs(G); return 0; }
运行结果:
(2)求出图G中出度最大的一个顶点,输出该顶点编号
#include "graph.h" //返回图G中编号为v的顶点的出度 int OutDegree(ALGraph *G,int v) { ArcNode *p; int n=0; p=G->adjlist[v].firstarc; while (p!=NULL) { n++; p=p->nextarc; } return n; } //输出图G中每个顶点的出度 void OutDs(ALGraph *G) { int i; for (i=0; i<G->n; i++) printf(" 顶点%d:%d\n",i,OutDegree(G,i)); } //输出图G中出度最大的一个顶点 void OutMaxDs(ALGraph *G) { int maxv=0,maxds=0,i,x; for (i=0; i<G->n; i++) { x=OutDegree(G,i); if (x>maxds) { maxds=x; maxv=i; } } printf("顶点%d,出度=%d\n",maxv,maxds); } int main() { ALGraph *G; int A[7][7]= { {0,1,1,1,0,0,0}, {0,0,0,0,1,0,0}, {0,0,0,0,1,1,0}, {0,0,0,0,0,0,1}, {0,0,0,0,0,0,0}, {0,0,0,1,1,0,1}, {0,1,0,0,0,0,0} }; ArrayToList(A[0], 7, G); printf("最大出度的顶点信息:"); OutMaxDs(G); return 0; }
运行结果:
(3)计算图G中出度为0的顶点数;
<span style="font-size:12px;"><span style="font-family:SimSun;">#include "graph.h" </span><span style="font-family:SimSun;color:#000000;">//返回图G中编号为v的顶点的出度 int OutDegree(ALGraph *G,int v) { ArcNode *p; int n=0; p=G->adjlist[v].firstarc; while (p!=NULL) { n++; p=p->nextarc; } return n; } //输出图G中每个顶点的出度 void OutDs(ALGraph *G) { int i; for (i=0; i<G->n; i++) printf(" 顶点%d:%d\n",i,OutDegree(G,i)); } //输出图G中出度为0的顶点数 void ZeroDs(ALGraph *G) { int i,x; for (i=0; i<G->n; i++) { x=OutDegree(G,i); if (x==0) printf("%2d",i); } printf("\n"); } int main() { ALGraph *G; int A[7][7]= { {0,1,1,1,0,0,0}, {0,0,0,0,1,0,0}, {0,0,0,0,1,1,0}, {0,0,0,0,0,0,1}, {0,0,0,0,0,0,0}, {0,0,0,1,1,0,1}, {0,1,0,0,0,0,0} }; ArrayToList(A[0], 7, G); printf("出度为0的顶点:"); ZeroDs(G); return 0; }</span></span>
运行结果:
#include "graph.h" //返回图G中编号为v的顶点的出度 int OutDegree(ALGraph *G,int v) { ArcNode *p; int n=0; p=G->adjlist[v].firstarc; while (p!=NULL) { n++; p=p->nextarc; } return n; } //输出图G中每个顶点的出度 void OutDs(ALGraph *G) { int i; for (i=0; i<G->n; i++) printf(" 顶点%d:%d\n",i,OutDegree(G,i)); } //输出图G中出度最大的一个顶点 void OutMaxDs(ALGraph *G) { int maxv=0,maxds=0,i,x; for (i=0; i<G->n; i++) { x=OutDegree(G,i); if (x>maxds) { maxds=x; maxv=i; } } printf("顶点%d,出度=%d\n",maxv,maxds); } //输出图G中出度为0的顶点数 void ZeroDs(ALGraph *G) { int i,x; for (i=0; i<G->n; i++) { x=OutDegree(G,i); if (x==0) printf("%2d",i); } printf("\n"); } //返回图G中是否存在边<i,j> bool Arc(ALGraph *G, int i,int j) { ArcNode *p; bool found = false; p=G->adjlist[i].firstarc; while (p!=NULL) { if(p->adjvex==j) { found = true; break; } p=p->nextarc; } return found; } int main() { ALGraph *G; int A[7][7]= { {0,1,1,1,0,0,0}, {0,0,0,0,1,0,0}, {0,0,0,0,1,1,0}, {0,0,0,0,0,0,1}, {0,0,0,0,0,0,0}, {0,0,0,1,1,0,1}, {0,1,0,0,0,0,0} }; ArrayToList(A[0], 7, G); printf("边<2,6>存在吗?"); if(Arc(G,2,6)) printf("是\n"); else printf("否\n"); printf("\n"); return 0; }