单源最短路径的算法求从任意指定的一个城市出发，到过所有其它城市的一条代价最小（距离最短）的路线

图的应用

• 1.题目描述
  已知中国12个城市及它们的经纬度：北京 北纬39.55 东经116.24 、兰州 北纬36.04 东经103.51 、广州 北纬23.08 东经113.14 、厦门 北纬24.27 东经118.06 、桂林 北纬25.17 东经110.17 、遵义 北纬27.42 东经106.55 、济南 北纬36.40 东经117.00 、郑州 北纬34.46 东经113.40 、哈尔滨 北纬45.44 东经126.36 、武汉 北纬30.35 东经114.17 、长沙 北纬28.12 东经112.59 、呼和浩特 北纬40.48 东经111.41。其中，北京至兰州，北京至桂林，北京至武汉，北京至长沙，桂林至广州，桂林至厦门，兰州至厦门，兰州至哈尔滨，广州至遵义，广州至哈尔滨，厦门至呼尔浩特，呼尔浩特至郑州，郑州至济南，遵义至长沙，遵义至武汉之间有交通线路，请用单源最短路径的算法求出从任意指定的一个城市出发，到过所有其它城市的一条代价最小（距离最短）的路线。
• 2 .代码如下

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define MAXSIZE 20
#define CITYNUMBER 12
#define ROADNUMBER 15

typedef struct node{
    int adjvex;
    float weight;
    struct node *next;
} edgenode;

typedef struct{
    char city_name[MAXSIZE];
    float location[2];
} cityinformation;

typedef struct {
    cityinformation data;
    edgenode*link;
} citynode;

typedef struct {
    int flag;
    int last_city;
    float distant;
} city;


Char cities[CITYNUMBER][MAXSIZE]={{“北京”},{“兰州”},{“广州”},{“厦门”},{“桂林”},{“遵义”},{“济南”},{“郑州”},{“哈尔滨”},{“武汉”},{“长沙”},{“呼和浩特”}
};
float location[12][2]={{39.55,116.24},{36.04,103.51},{23.08,113.14},{24.27,118.06},{25.17,110.17},{27.42,106.55},{36.40,117.00},{34.46,113.40},{45.44,126.36},{30.35,114.17},{28.12,112.59},{40.48,111.41}
};
int connection[15][2]={{0,1},{0,4},{0,9},{0,10},{4,2},{4,3},{1,3},{1,8},{2,5},{2,8},{3,11},{11,7},{7,6},{5,10},{5,9}
};


double CountWeight(cityinformation ,cityinformation);
void CreateAdjList(citynode G[]);
void FindShortestWay(citynode G[]);
void __OutputResult__(city [],citynode []);


int main(int argc, const char * argv[]) {
    citynode G[CITYNUMBER];
    CreateAdjList(G);
    FindShortestWay(G);
    return 0;
}

void CreateAdjList(citynode G[]){
    int start,end;
    edgenode *tmp;
    for (int I=0;i<CITYNUMBER;i++){
        strcpy(G[I].data.city_name,cities[I]);
        G[I].data.location[0]=location[I][0];
        G[I].data.location[1]=location[I][1];
        G[I].link=NULL;
    }
    for (int j=0; j<ROADNUMBER; j++){
       
        start=connection[j][0];
        end=connection[j][1];
        int weight=CountWeight(G[start].data, G[end].data);
        tmp=(edgenode*)malloc(sizeof(edgenode));
        tmp->adjvex=end;
        tmp->weight=weight;
        tmp->next=G[start].link;
        G[start].link=tmp;
        
        tmp=(edgenode*)malloc(sizeof(edgenode));
        tmp->adjvex=start;
        tmp->weight=weight;
        tmp->next=G[end].link;
        G[end].link=tmp;
        
    }
}

double CountWeight(cityinformation a,cityinformation b){
    double result;
    result =sqrt((a.location[0]-b.location[0])*(a.location[0]-b.location[0])+(a.location[1]-b.location[1])*(a.location[1]-b.location[1]));
    return result;
}


void FindShortestWay(citynode G[]){
    city tmp [CITYNUMBER];
    char start_cityname[MAXSIZE];
    int start_city_id;
    printf(“请输入起点城市名:”);
    scanf(“%s”,start_cityname);
    for (int i=0;i<CITYNUMBER;i++){
        if(!strcmp(G[I].data.city_name, start_cityname)){
            start_city_id=I;
            tmp[I].flag=1;
            tmp[I].distant=0;
            tmp[I].last_city=-1;
        }else{
        tmp[I].flag=0;
        tmp[I].distant=9999;
        }
    }
    edgenode*a = G[start_city_id].link;
    while (a){
        tmp[a->adjvex].distant = a->weight;
        tmp[a->adjvex].last_city = start_city_id;
        a=a->next;
    }//初始化完成
    for(int I=0;i<CITYNUMBER-1;i++){
        int min=10000,next_shortest_city_id;
        for (int j=0; j<CITYNUMBER; j++) {
            if((!tmp[j].flag)&&(tmp[j].distant<min)){
                min = tmp[j].distant;
                next_shortest_city_id = j;
            }
        }
        tmp[next_shortest_city_id].flag = 1;
        edgenode*a = G[next_shortest_city_id].link;
        while (a) {
            if(!tmp[a->adjvex].flag && tmp[a->adjvex].distant>tmp[next_shortest_city_id].distant+a->weight){
                tmp[a->adjvex].distant = tmp[next_shortest_city_id].distant+a->weight;
                tmp[a->adjvex].last_city = next_shortest_city_id;
            }
            a=a->next;
        }
    }
    __OutputResult__(tmp, G);
}

void __OutputResult__(city a[],citynode b[]){
    int tmp[CITYNUMBER];
    for(int i=0;i<CITYNUMBER;i++){
        int count = 0;
        int j=I;
        while (a[j].last_city!=-1) {
            tmp[count++]=j;
            j=a[j].last_city;
        }
        tmp[count]=j;
        int k;
        for(k=count;k>0;k—){
            printf(“%s->”,b[tmp[k]].data.city_name);
        }
        if(count)
            printf(“%s\n”,b[tmp[k]].data.city_name);
    }
}

*3.运行结果