Dijkstra最短路径算法实现代码

Dijkstra的最短路径算法是基于前驱顶点的最短路径计算的，整体上来讲还是比较简单的，下面是代码：

void shortestpath( const std::vector <std::vector< short> >& paths, int from, std::vector< short>& path){
    std:: vector<bool> flags(paths.size(), false);
    std:: vector<short> distance(paths.size(), 0);
    path.resize(paths.size(), 0);

    for(size_t i = 0; i != paths.size(); ++i){
        distance[i] = paths[from][i];
    }

    flags[from] = 1;

    int min, pos;
    for(size_t i = 1; i != paths.size(); ++i){
        pos = -1;
        min = std:: numeric_limits<short>::max();
        for(size_t j = 0; j != paths.size(); ++j){
            if(!flags[j] && distance[j] < min){
                min = distance[j];
                pos = j;
            }
        }

        if(pos == -1)
            break;

        flags[pos] = true;

        for(size_t j = 0; j != paths.size(); ++j){
            if(!flags[j] && paths[pos][j] != 0
                && paths[pos][j] < std::numeric_limits <int>:: max()
                && min+paths[pos][j] < distance[j]){
                distance[j] = min + paths[pos][j];
                path[j] = pos;
            }
        }
    }

    for(size_t i = 0; i != distance.size(); ++i){
        std::cout << distance[i] << " " << std::flush;
    }
    std::cout << std:: endl;
}

int main(){
    std::cout << "请输入顶点数：" << std::flush;
    int sum; std::cin >> sum;
    std:: vector<std::vector <short> > paths;
    for(int i = 0; i != sum; ++i){
        paths.push_back(std:: vector<short>(sum, std::numeric_limits< short>::max()));
        paths[i][i] = 0;
    }

    std::cout << "请输入边数：" << std::flush;
    std::cin >> sum;

    int vi, vj, weight;
    for(int i = 0; i != sum; ++i){
        std::cin >> vi >> vj >> weight;
        paths[vi][vj] = weight;
        paths[vj][vi] = weight;
    }

    std:: vector<short> path;
    shortestpath(paths, 0, path);

    std::cout << "最近路径结果为：" << std::flush;
    for(size_t i = 0; i != path.size(); ++i){
        std::cout << path[i] << " " << std::flush;
    }
    std::cout << std:: endl;

    return 0;
}
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