构建邻接矩阵有向图并进行最短路径算法和深度分析

Question

构建邻接矩阵有向图并进行最短路径算法和深度分析
这是我已经写的一部分代码
还有报错：
0x00007FF75D966847 处(位于 第一次实习.exe 中)引发的异常: 0xC0000005: 写入位置 0x000000CEF2400000 时发生访问冲突。
该怎么修改，并完善代码

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <string>
using namespace std;
 
 
//结构体
struct City {   //图的顶点
    string country, city;
    float latitude, longitude;//经纬度
};
struct Route {
    string start_city, end_city, transport, other_info;//起点，终点，交通工具，其他信息
    float time, cost;//时间，花费
};
 
struct Graph {
    Route arcs[199][199];//二维数组
    City vexs[199];//顶点数据信息
    int vexnum, arcnum;//顶点数和弧数
};
 
 
int LocateVex(string &G,City* c) {
    int i = 0;
    for (; i < 199; i++)
    {
        if (G == c[i].city) {
            break;
        }
        else {
            return -1;
        }
    }
    return i;
    
}
 
 
int main()
{
    //读取 cities 表
    FILE* fp;
    fp = fopen("C:\\Users\\张宇航\\Desktop\\集中实习\\cities.csv", "r");
    if (!fp) {
        printf("ERROR!!!");
        exit(0);
    }
    City city[199];
    char ch;
    float f;
    int i = 0, j = 0;
 
    while (!feof(fp))
    {
        //存储国家名
        ch = fgetc(fp);
        for (; ch != ','; ch = fgetc(fp))
        {
            city[i].country += ch;
        }
 
        //存储城市名
        ch = fgetc(fp);
        for (; ch != ','; ch = fgetc(fp))
        {
            city[i].city += ch;
        }
 
        //存储经纬度
        fscanf(fp, "%f,", &f);
        city[i].latitude = f;
        fscanf(fp, "%f\n", &f);
        city[i].longitude = f;
        i++;
    }
    fclose(fp);
    
    //读取 routes 表
    FILE* fp1;
    fp1 = fopen("C:\\Users\\张宇航\\Desktop\\集中实习\\routes.csv", "rt");
    if (!fp1) {
        printf("ERROR!!!");
        exit(0);
    }
    Route route[1975];
    i = 0;
 
    
    while (!feof(fp1))
    {
        //存储起点城市
        ch = fgetc(fp1);
        for (; ch != ','; ch = fgetc(fp1))
        {
            route[i].start_city += ch;
        }
 
        //存储终点城市
        ch = fgetc(fp1);
        for (; ch != ','; ch = fgetc(fp1))
        {
            route[i].end_city += ch;
        }
 
        //存储交通工具
        ch = fgetc(fp1);
        for (; ch != ','; ch = fgetc(fp1))
        {
            route[i].transport += ch;
        }
 
        //存储时间和花费
        fscanf(fp1, "%f,", &route[i].time);
        fscanf(fp1, "%f,", &route[i].cost);
 
        //存储其他信息
        ch = fgetc(fp1);
        for (; ch != '\n'; ch = fgetc(fp1))
        {
            route[i].other_info += ch;
        }
        
        i++;
    }
 
    //建立邻接矩阵
    Graph graph;
    graph.arcnum = 1975;
    graph.vexnum = 199;
    
    //构建图
    //依次录入顶点的数据
    for (int i = 0; i < (graph.vexnum); i++)
    {
        graph.vexs[i] = city[i];
    }
    //初始化二维矩阵
    for (int i = 0; i < graph.vexnum; i++)
    {
        for (int j = 0; j < graph.vexnum; j++)
        {
            if (i == j) {
                graph.arcs[i][j].cost = 0;
                graph.arcs[i][j].time = 0;
            }
            else {
                graph.arcs[i][j].cost = 1;
                graph.arcs[i][j].time = 1;
            }
        }
    }
    //添加弧数据
    for (int i = 0; i < graph.arcnum; i++)
    {
        int sta = LocateVex(route[i].start_city, city);
        int end = LocateVex(route[i].end_city, city);
        graph.arcs[sta][end].cost = route[i].cost;
        graph.arcs[sta][end].time = route[i].time;
    }
    return 0;
}
 

Answer 1

• 你可以看下这个问题的回答https://ask.csdn.net/questions/7752734