图形结构源码(do it myself)

1、Kruskal算法

#include <stdio.h>#include <stdlib.h>struct list{int vertex1;int vertex2;int weight;struct list *next;};typedef struct list *edge;#define vertexmax 6            //从下标1存入5点(1~5)#define edgenum 10             //边数int visit[vertexmax];          //标记某顶点是否被访问int Edges[10][3] =                /* 输入数据 */{ {1,2,7}, {1,3,6}, {1,4,5}, {1,5,12}, {2,3,14},{2,4,8}, {2,5,8}, {3,4,3}, {3,5,9}, {4,5,2} };edge creategraph(edge head)            //按边权重排序各边，生成链表{int i;edge New,pointer;for(i=1;i<edgenum;i++){New=(edge)malloc(sizeof(struct list));pointer=head;New->vertex1=Edges[i][0];New->vertex2=Edges[i][1];New->weight=Edges[i][2];New->next=NULL;while(pointer!=NULL){if(New->weight<head->weight){New->next=head;head=New;break;}if(New->weight>=pointer->weight && pointer->next==NULL){pointer->next=New;break;}if(New->weight>=pointer->weight && New->weight<pointer->next->weight){New->next=pointer->next;pointer->next=New;break;}pointer=pointer->next;}}return head;}void print(edge head)            //打印链表{edge pointer=head;while(pointer!=NULL){printf("(%d,%d)[%d]",pointer->vertex1,pointer->vertex2,pointer->weight);pointer=pointer->next;}printf("\n");}void Krusral(edge head)           //krusral算法{int minedgenum=0;             int weight=0;edge pointer=head;while(pointer!=NULL){if(visit[pointer->vertex1]==0 || visit[pointer->vertex2]==0)      //当边的两端点有一个未访问时{printf("(%d%d)[%d] ",pointer->vertex1,pointer->vertex2,pointer->weight);weight=weight+pointer->weight;visit[pointer->vertex1]=1;visit[pointer->vertex2]=1;minedgenum++;}if(minedgenum==vertexmax-2)         //生产树的变数等于(顶点数-1)时，结束{printf("\nthe sum of the weight is %d\n",weight);break;}pointer=pointer->next;}}void main(){edge head;int i;for(i=0;i<vertexmax;i++){visit[i]=0;}head=(edge)malloc(sizeof(struct list));if(head!=NULL){head->vertex1=1;head->vertex2=2;head->weight=7;head->next=NULL;head=creategraph(head);print(head);Krusral(head);}system("pause");}

2、Prims算法

#include <stdio.h>#include <stdlib.h>struct list{int vertex1;int vertex2;int weight;int flag;struct list *next;};struct node{int index;struct list *next;};typedef struct list *edge;typedef struct node *Node;#define vertexmax 6            //从下标1存入5点(1~5)#define edgenum 10             //边数struct node head[vertexmax];int visit[vertexmax];          //标记某顶点是否被访问int Edges[10][3] =                /* 输入数据 */{ {1,2,7}, {1,3,6}, {1,4,5}, {1,5,12}, {2,3,14},{2,4,8}, {2,5,8}, {3,4,3}, {3,5,9}, {4,5,2} };void creategraph(Node head)            //按边权重排序各边，生成链表{int i,j; edge pointer,New;for(i=1;i<vertexmax;i++){for(j=0;j<edgenum;j++){pointer=head[i].next;if(Edges[j][0]==i){New=(edge)malloc(sizeof(struct list));New->flag=0;New->vertex1=Edges[j][0];New->vertex2=Edges[j][1];New->weight=Edges[j][2];New->next=NULL;/*while(pointer!=NULL){pointer=pointer->next;}*/New->next=pointer;head[i].next=New;}}}}void print(Node head)            //打印链表{edge pointer;int i;for(i=1;i<vertexmax;i++){pointer=head[i].next;while(pointer!=NULL){printf("(%d,%d)",pointer->vertex1,pointer->vertex2);pointer=pointer->next;}if(i==vertexmax-1)break;printf("\n");}}void prims(Node head,int index)           //krusral算法{int i;int vertex2;int minedgenum=1;edge pointer,minedge;visit[index]=1;while(minedgenum<=vertexmax-2){minedge=(edge)malloc(sizeof(struct list));minedge->weight=999;for(i=1;i<vertexmax;i++){pointer=head[i].next;if(pointer==NULL)continue;else{do{if((pointer->flag==0) && (minedge->weight > pointer->weight) //visit[pointer->v1和v2]只能有一个被访问过&& (visit[pointer->vertex1]==1^visit[pointer->vertex2]==1) &&(pointer->vertex1==i || pointer->vertex2==i)){minedge=pointer;vertex2=minedge->vertex2;}pointer=pointer->next;}while(pointer!=NULL);}}printf("(%d,%d) ",minedge->vertex1,minedge->vertex2);visit[minedge->vertex1]=1;visit[minedge->vertex2]=1;minedgenum++;minedge->flag=1;}printf("\n");}void main(){int i;for(i=0;i<vertexmax;i++){visit[i]=0;}for(i=0;i<vertexmax;i++){head[i].index=i;head[i].next=NULL;}creategraph(head);print(head);printf("the min tree is counstructed by the edges: ");prims(head,1);system("pause");}