如何在C++路径寻优算法文件中嵌入样本数据直接运行

问题描述 #

现有一段用于寻找最短且盈利最高配对路径的C++算法代码，原本通过命令行参数读取外部图数据文件。需要将提供的样本图数据直接嵌入到.cpp文件中，无需依赖外部文件即可运行并得到结果，不清楚如何构建对应数据结构。

解决方案 #

核心思路是用内存字符串流（istringstream）模拟文件输入，替代原来的ifstream读取外部文件，同时将命令行参数硬编码为固定值。这样无需修改readGraph和bellmanFord等核心函数的逻辑，只需调整数据输入和参数传递部分。

修改步骤 #

1. 添加必要头文件：引入<sstream>以使用字符串流
2. 嵌入样本数据：将样本数据格式化为多行字符串，保持与原文件一致的结构
3. 替换文件输入流：用istringstream替代ifstream，读取内存中的数据
4. 硬编码命令行参数：将原本通过argv获取的输入输出路径、源节点、目标节点、交易成本等参数改为固定值

修改后的完整代码 #

#include "readGraph.hpp"
#include "ShortestPath.hpp"
#include <limits>
#include <fstream>
#include <string>
#include <iostream>
#include <cmath>
#include <chrono>
#include <sstream>  // 新增头文件
using namespace std;

int main() {  // 去掉命令行参数，改为无参main
    // -------------------------- 嵌入样本数据（样本2） --------------------------
    const string graphData = R"(6 30

GBP

EUR

JPY

CHF

USD

GLD

0 1 1.4599 GBP->EUR

0 2 189.05 GBP->JPY

0 3 2.1904 GBP->CHF

0 4 1.5714 GBP->USD

0 5 0.004816 GBP->GLD

1 0 0.684978 EUR->GBP

1 2 129.52 EUR->JPY

1 3 1.4978 EUR->CHF

1 4 1.0752 EUR->USD

1 5 0.003295 EUR->GLD

2 0 0.00528961 JPY->GBP

2 1 0.00772082 JPY->EUR

2 3 0.011574 JPY->CHF

2 4 0.008309 JPY->USD

2 5 0.0000255 JPY->GLD

3 0 0.456538 CHF->GBP

3 1 0.667646 CHF->EUR

3 2 86.4006 CHF->JPY

3 4 0.7182 CHF->USD

3 5 0.002201 CHF->GLD

4 0 0.636375 USD->GBP

4 1 0.93006 USD->EUR

4 2 120.351 USD->JPY

4 3 1.39237 USD->CHF

4 5 0.003065 USD->GLD

5 0 207.641 GLD->GBP

5 1 303.49 GLD->EUR

5 2 39215.7 GLD->JPY

5 3 454.339 GLD->CHF

5 4 326.264 GLD->USD
)";
    // -------------------------------------------------------------------------

    // 用字符串流模拟文件输入
    istringstream fin(graphData);
    // 硬编码输出文件路径（或改为cout直接输出到控制台）
    ofstream fout("output.txt");

    // 硬编码交易成本、源节点、目标节点
    double cost = 0.0;  // 可根据需求修改
    string sourceV = "GBP";
    string destV = "EUR";

    int** edgeL;
    double* weights;
    int numEdges;
    double totalDist = 0.0;  // 初始化totalDist，避免未定义行为
    string* e; //eLabels
    string* v; //vLabels

    /* edgeList version of readGraph */
    int numVertices = readGraph(fin, edgeL, weights, numEdges, v, e);

    /* Checking input source and destination vertices and assigning their corresponding integer values */ 
    int source = -1, dest = -1;
    for(int i = 0; i < numVertices; i ++){
        if(sourceV == v[i]){
            source = i;
        }
        if(destV == v[i]){
            dest = i; 
        }
    }
    // 检查源/目标节点是否存在
    if (source == -1 || dest == -1) {
        cerr << "Source or destination vertex not found!" << endl;
        return 1;
    }
    
    int* prev;
    double* dist;
    int* path;
    int* cycle;

    /*Prepping the weights for bellmanFord by taking setting w[i] = -log(temp*(1-cost)) */
    for(int i = 0; i < numEdges; i ++){
        double temp = weights[i];
        weights[i] = -1*log(temp*(1-cost));
        // cout << weights[i] << endl; // 可选：注释掉减少输出
    }
    
    /* Setting timers and invoking bellmanFord to catch arbitrage opportunity */
    auto t1 = std::chrono::system_clock::now();
    int x = bellmanFord(edgeL, weights, numVertices, numEdges, source, dist, prev);
    auto t2 = std::chrono::system_clock::now();
    auto dur = t2-t1;
    auto durns = std::chrono::duration_cast<std::chrono::microseconds>(dur);
    double elapsed = durns.count();
    
   
    /** If there is a negative loop, there is an arbitrage opportunity.
     *  A path is derived using the appropriate getCycle method. 
     *  The path is printed to fout and meta data printed to the terminal.
    */
    if(x != -1){
        int cycleSize = getCycle(x, prev, numVertices, cycle);
        fout << numVertices << " " << cycleSize << endl;
        
        for(int i = 0; i < numVertices; i ++){
            fout << v[i] << endl;
        }
        
        int idx = 0;
        for(int i = 0; i < cycleSize; i ++){
            for(int j = 0; j < numEdges; j ++){
                if(edgeL[j][0] == cycle[i] && edgeL[j][1] == cycle[i+1]){
                    totalDist += weights[j];
                    fout << cycle[i] << " " << cycle[i+1] << " " << exp(-1*weights[j]) << " " << e[j] << endl;
                    idx = j;
                }
            }
        }
        
        cout << "Arbitrage Opportunity Detected!" << endl;
        cout << "Effective Exchange Rate: " << exp(-1*totalDist) << endl;
        cout << "Runtime: " << elapsed << " Microseconds" << endl;

        delete[] cycle;
        for (int i=0; i<numEdges; i++){
            delete[] edgeL[i];
        }
    }

    /** If there is no negative cycle. 
     *  The path is simply printed to fout and meta data printed to terminal. 
     */
    else{  // 简化判断，直接else即可

        int pathSize = getPath(source, dest, prev, path);
        fout << numVertices << " " << pathSize - 1 << endl;
        
        for(int i = 0; i < numVertices; i++){
            fout << v[i] << endl;
        }
        
        int temp;
        
        for(int i = 0; i < pathSize-1; i++){
            temp = path[i];
            for(int j = 0; j < numEdges; j ++){
                if(edgeL[j][0] == path[i] && edgeL[j][1] == path[i+1]){
                    totalDist += weights[j];
                    fout << temp << " " << path[i+1] << " " << exp(-1*weights[j]) << " " << e[j] << endl;
                }
            }
        }
        cout << "Effective Exchange Rate: " << exp(-1*totalDist) << endl;
        cout << "Runtime: " << elapsed << " Microseconds" << endl;

        delete[] path;
        for (int i=0; i<numEdges; i++){
            delete[] edgeL[i];
        }    
    }

    delete[] dist;
    delete[] prev;
    delete[] e;
    delete[] v;
    delete[] weights;
    delete[] edgeL;

    return 0;
}

关键说明 #

1. 样本数据嵌入：使用C++11及以上支持的原始字符串字面量（R"(...)"），可以直接保留样本数据的换行和空格，避免转义字符的麻烦。如果使用旧标准，可以用"\n"替换换行。
2. 字符串流替代文件流：istringstream会将字符串内容模拟成文件输入流，完全兼容原有的readGraph函数，无需修改该函数的实现。
3. 参数硬编码：将cost、sourceV、destV等参数改为固定值，同时去掉main函数的命令行参数，简化运行流程。
4. 修复潜在问题：初始化totalDist为0，避免未定义行为；增加源/目标节点存在性检查，提升代码健壮性。

切换样本数据 #

如果需要使用样本1，只需替换graphData的内容为：

const string graphData = R"( 6 1


GBP

EUR

JPY

CHF

USD

GLD

0 1 1.45543 GBP->EUR
)";

内容的提问来源于stack exchange，提问作者Udbdjfj