1、是否有简单路径?
问题:假设图G采用邻接表存储,设计一个算法,判断顶点u到v是否有简单路径。
#include <stdio.h>#include <malloc.h>#include "graph.h"int visited[MAXV]; void ExistPath(ALGraph *G,int u,int v, bool &has){ int w; ArcNode *p; visited[u]=1; if(u==v) { has=true; return; } p=G->adjlist[u].firstarc; while (p!=NULL) { w=p->adjvex; if (visited[w]==0) ExistPath(G,w,v,has); p=p->nextarc; }}void HasPath(ALGraph *G,int u,int v){ int i; bool flag = false; for (i=0; i<G->n; i++) visited[i]=0; ExistPath(G,u,v,flag); printf(" 从 %d 到 %d ", u, v); if(flag) printf("有简单路径\n"); else printf("无简单路径\n");}int main(){ ALGraph *G; int A[5][5]= { {0,0,0,0,0}, {0,0,1,0,0}, {0,0,0,1,1}, {0,0,0,0,0}, {1,0,0,1,0}, }; ArrayToList(A[0], 5, G); HasPath(G, 1, 0); HasPath(G, 4, 1); return 0;}
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附:测试图结构及存储
2、输出简单路径
问题:假设图G采用邻接表存储,设计一个算法输出图G中从顶点u到v的一条简单路径(假设图G中从顶点u到v至少有一条简单路径)。
#include <stdio.h>#include <malloc.h>#include "graph.h"int visited[MAXV]; void FindAPath(ALGraph *G,int u,int v,int path[],int d){ int w,i; ArcNode *p; visited[u]=1; d++; path[d]=u; if (u==v) { printf("一条简单路径为:"); for (i=0; i<=d; i++) printf("%d ",path[i]); printf("\n"); return; } p=G->adjlist[u].firstarc; while (p!=NULL) { w=p->adjvex; if (visited[w]==0) FindAPath(G,w,v,path,d); p=p->nextarc; }}void APath(ALGraph *G,int u,int v){ int i; int path[MAXV]; for (i=0; i<G->n; i++) visited[i]=0; FindAPath(G,u,v,path,-1); }int main(){ ALGraph *G; int A[5][5]= { {0,0,0,0,0}, {0,0,1,0,0}, {0,0,0,1,1}, {0,0,0,0,0}, {1,0,0,1,0}, }; ArrayToList(A[0], 5, G); APath(G, 1, 0); APath(G, 4, 1); return 0;}
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3、输出所有路径
问题:输出从顶点u到v的所有简单路径。
#include <stdio.h>#include <malloc.h>#include "graph.h"int visited[MAXV]; void FindPaths(ALGraph *G,int u,int v,int path[],int d){ int w,i; ArcNode *p; visited[u]=1; d++; path[d]=u; if (u==v && d>1) { printf(" "); for (i=0; i<=d; i++) printf("%d ",path[i]); printf("\n"); } p=G->adjlist[u].firstarc; while(p!=NULL) { w=p->adjvex; if (visited[w]==0) FindPaths(G,w,v,path,d); p=p->nextarc; } visited[u]=0; }void DispPaths(ALGraph *G,int u,int v){ int i; int path[MAXV]; for (i=0; i<G->n; i++) visited[i]=0; printf("从%d到%d的所有路径:\n",u,v); FindPaths(G,u,v,path,-1); printf("\n");}int main(){ ALGraph *G; int A[5][5]= { {0,1,0,1,0}, {1,0,1,0,0}, {0,1,0,1,1}, {1,0,1,0,1}, {0,0,1,1,0} }; ArrayToList(A[0], 5, G); DispPaths(G, 1, 4); return 0;}
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附:测试用的图结构、存储结构、运行结果
4、输出一些简单回路
问题:输出图G中从顶点u到v的长度为s的所有简单路径。
#include <stdio.h>#include <malloc.h>#include "graph.h"int visited[MAXV]; void SomePaths(ALGraph *G,int u,int v,int s, int path[],int d){ int w,i; ArcNode *p; visited[u]=1; d++; path[d]=u; if (u==v && d==s) { printf(" "); for (i=0; i<=d; i++) printf("%d ",path[i]); printf("\n"); } p=G->adjlist[u].firstarc; while(p!=NULL) { w=p->adjvex; if (visited[w]==0) SomePaths(G,w,v,s,path,d); p=p->nextarc; } visited[u]=0; }void DispSomePaths(ALGraph *G,int u,int v, int s){ int i; int path[MAXV]; for (i=0; i<G->n; i++) visited[i]=0; printf("从%d到%d长为%d的路径:\n",u,v,s); SomePaths(G,u,v,s,path,-1); printf("\n");}int main(){ ALGraph *G; int A[5][5]= { {0,1,0,1,0}, {1,0,1,0,0}, {0,1,0,1,1}, {1,0,1,0,1}, {0,0,1,1,0} }; ArrayToList(A[0], 5, G); DispSomePaths(G, 1, 4, 3); return 0;}
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附:测试的图结构、运行结果
5、输出通过一个节点的所有简单回路
问题:求图中通过某顶点k的所有简单回路(若存在)
#include <stdio.h>#include <malloc.h>#include "graph.h"int visited[MAXV]; void DFSPath(ALGraph *G,int u,int v,int path[],int d){ int w,i; ArcNode *p; visited[u]=1; d++; path[d]=u; p=G->adjlist[u].firstarc; while (p!=NULL) { w=p->adjvex; if (w==v && d>0) { printf(" "); for (i=0; i<=d; i++) printf("%d ",path[i]); printf("%d \n",v); } if (visited[w]==0) DFSPath(G,w,v,path,d); p=p->nextarc; } visited[u]=0; }void FindCyclePath(ALGraph *G,int k){ int path[MAXV],i; for (i=0; i<G->n; i++) visited[i]=0; printf("经过顶点%d的所有回路\n",k); DFSPath(G,k,k,path,-1); printf("\n");}int main(){ ALGraph *G; int A[5][5]= { {0,1,1,0,0}, {0,0,1,0,0}, {0,0,0,1,1}, {0,0,0,0,1}, {1,0,0,0,0} }; ArrayToList(A[0], 5, G); FindCyclePath(G, 0); return 0;}
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附:测试用图结构、输出结果
1、最短路径
问题:求不带权连通图G中从顶点u到顶点v的一条最短路径。
#include <stdio.h>#include <malloc.h>#include "graph.h"typedef struct{ int data; int parent; } QUERE; void ShortPath(ALGraph *G,int u,int v){ ArcNode *p; int w,i; QUERE qu[MAXV]; int front=-1,rear=-1; int visited[MAXV]; for (i=0; i<G->n; i++) visited[i]=0; rear++; qu[rear].data=u; qu[rear].parent=-1; visited[u]=1; while (front!=rear) { front++; w=qu[front].data; if (w==v) { i=front; while (qu[i].parent!=-1) { printf("%2d ",qu[i].data); i=qu[i].parent; } printf("%2d\n",qu[i].data); break; } p=G->adjlist[w].firstarc; while (p!=NULL) { if (visited[p->adjvex]==0) { visited[p->adjvex]=1; rear++; qu[rear].data=p->adjvex; qu[rear].parent=front; } p=p->nextarc; } }}int main(){ ALGraph *G; int A[9][9]= { {0,1,1,0,0,0,0,0,0}, {0,0,0,1,1,0,0,0,0}, {0,0,0,0,1,1,0,0,0}, {0,0,0,0,0,0,1,0,0}, {0,0,0,0,0,1,1,0,0}, {0,0,0,0,0,0,0,1,0}, {0,0,0,0,0,0,0,1,1}, {0,0,0,0,0,0,0,0,1}, {0,0,0,0,0,0,0,0,0} }; ArrayToList(A[0], 9, G); ShortPath(G,0,7); return 0;}
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附:测试用图结构
2、最远顶点
问题:求不带权连通图G中,距离顶点v最远的顶点k
#include <stdio.h>#include <malloc.h>#include "graph.h"int Maxdist(ALGraph *G,int v){ ArcNode *p; int i,j,k; int Qu[MAXV]; int visited[MAXV]; int front=0,rear=0; for (i=0; i<G->n; i++) visited[i]=0; rear++; Qu[rear]=v; visited[v]=1; while (rear!=front) { front=(front+1)%MAXV; k=Qu[front]; p=G->adjlist[k].firstarc; while (p!=NULL) { j=p->adjvex; if (visited[j]==0) { visited[j]=1; rear=(rear+1)%MAXV; Qu[rear]=j; } p=p->nextarc; } } return k;}int main(){ ALGraph *G; int A[9][9]= { {0,1,1,0,0,0,0,0,0}, {0,0,0,1,1,0,0,0,0}, {0,0,0,0,1,1,0,0,0}, {0,0,0,0,0,0,1,0,0}, {0,0,0,0,0,1,1,0,0}, {0,0,0,0,0,0,0,1,0}, {0,0,0,0,0,0,0,1,1}, {0,0,0,0,0,0,0,0,1}, {0,0,0,0,0,0,0,0,0} }; ArrayToList(A[0], 9, G); printf("离顶点0最远的顶点:%d",Maxdist(G,0)); return 0;}
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附:测试用图结构