c++ - 非递归 Kosaraju 的两遍算法实现永远在大型数据集上执行

Question

• 我为已过截止日期的作业编写了此代码。

• 此实现完全适用于各种较小的测试用例，并在图表中显示 5 个最大的强连通组件的大小。

• 但是当我在大约 875714 个顶点的分配数据集上运行它时，它似乎永远执行。 (60 分钟后甚至没有从第一个 DFS 通过)

• 我使用了 DFS 例程的非递归堆栈实现，因为我听说大量的顶点导致递归堆栈溢出问题。

• 如果有人能指出这段代码中的什么使得它在大型数据集上以这种方式表现，那将非常有帮助。

• 输入文件由图中的边列表组成。一条边/一条线。

(例如):

1 2

2 3

3 1

3 4

5 4

Download link for the Large graph test case zip file

Link to my program file

代码如下:

//宏定义和全局变量

#define N 875714
#define all(a) (a).begin(), (a).end()
#define tr(c,i) for(typeof((c).begin()) i = (c).begin(); i != (c).end(); i++)

vi v(N), ft, size;

//非递归DFS算法

void DFS(vvi g, int s, int flag)
{
stack<int> stk;
stk.push(s);
v[s] = 1;

int jumpOut, count;
vi::iterator i;

if(flag == 2)
     count = 1;

while(!stk.empty())
{
i = g[stk.top()].begin();
jumpOut = 0;

for(; i != g[stk.top()].end(); i++)
{
    if(v[*i] != 1)
    {
        stk.push(*i);
        v[*i] = 1;

        if(flag == 2) //Count the SCC size
            count++;

        jumpOut = 1; //Jump to the while loop's beginning
        break;
    }
 }

 if(flag == 1 && jumpOut == 0) //Record the finishing time order of vertices
    ft.push_back(stk.top());

 if(jumpOut == 0)
      stk.pop();
}

if(flag == 2)
    size.push_back(count); //Store the SCC size
}

//2 pass Kosaraju 算法

void kosaraju(vvi g, vvi gr)
{
cout<<"\nInside pass 1\n";

for(int i = N - 1; i >= 0; i--)
    if(v[i] != 1)
        DFS(gr, i, 1);

cout<<"\nPass 1 completed\n";

fill(all(v), 0);

cout<<"\nInside pass 2\n";

for(int i = N - 1; i >= 0; i--)
    if(v[ ft[i] ] != 1)
        DFS(g, ft[i], 2);

cout<<"\nPass 2 completed\n";
}

.

int main()
{
vvi g(N), gr(N);
ifstream file("/home/tauseef/Desktop/DAA/SCC.txt");
int first, second;
string line;

while(getline(file,line,'\n')) //Reading from file
{
    stringstream ss(line);
    ss >> first;
    ss >> second;
    if(first == second) //Eliminating self loops
        continue;

    g[first-1].push_back(second-1); //Creating G & Grev
    gr[second-1].push_back(first-1);
}

cout<<"\nfile read successfully\n";

kosaraju(g, gr);

cout<<"\nFinishing order is: ";
tr(ft, j)
    cout<<*j+1<<" ";
cout<<"\n";

sort(size.rbegin(), size.rend()); //Sorting the SCC sizes in descending order

cout<<"\nThe largest 5 SCCs are: ";
tr(size, j)
    cout<<*j<<" ";
cout<<"\n";

file.close();
}

Answer 1

您可以应用多项改进:
1- cin 对于大的输入没有那么快 scanf:因为你的输入文件很大你最好使用 scanf读取您的数据。
2- 按值将大数据传递给函数不是一个好主意:您的代码中有两个巨大的图形，您将它们按值传递给函数。这需要很多时间，因为每次你都在复制数据。
3- 无需使用iterator 来遍历vector:因为您正在使用vector 并且您通过 [] 运算符可以随机访问它，无需使用 iterator 来访问数据。
4- 你的 DFS 效率不高:这是最重要的一个。每次程序转到 while 的开头并检查 stack 顶部元素的邻接列表时，您从头开始检查元素。这使得算法非常低效，因为你要一遍又一遍地检查一些东西。您可以简单地存储您检查了多少个 child ，当您返回到该元素时，您从下一个元素开始，而不是从 start 开始。

#include<iostream>
#include<vector>
#include<stack>
#include<algorithm>
#include<fstream>
#include<string>
#include<sstream>
using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;

#define N 875714
#define sz(a) int((a).size())
#define all(a) (a).begin(), (a).end()
#define tr(c,i) for(typeof((c).begin()) i = (c).begin(); i != (c).end(); i++)

vi v(N), ft, size;
vi childsVisited(N);

void DFS(vvi &g, int s, int flag)
{
    stack<int> stk;
    stk.push(s);
    v[s] = 1;

    int jumpOut, count;

    if(flag == 2)
        count = 1;
    int counter = 0;
    while(!stk.empty())
    {
        jumpOut = 0;
        int cur = stk.top();
        for ( ;childsVisited[cur] < g[cur].size(); ++childsVisited[cur] )
        //for ( int i=0; i< g[cur].size(); ++i )
        //for(; i != g[stk.top()].end(); i++)
        {
            int i = childsVisited[cur];
            int next = g[cur][i];
            if(v[next] != 1)
            {
                stk.push(next);
                v[next] = 1;
                if(flag == 2) //Count the SCC size
                    count++;

                jumpOut = 1; //Jump to the while loop's beginning
                break;
            }
        }

        if(flag == 1 && jumpOut == 0) //Record the finishing time order of vertices
            ft.push_back(stk.top());

        if(jumpOut == 0)
            stk.pop();
    }

    if(flag == 2)
        size.push_back(count); //Store the SCC size
}

void kosaraju(vvi &g, vvi &gr)
{
    cout<<"\nInside pass 1\n";

    for(int i = N - 1; i >= 0; i--)
        if(v[i] != 1)
            DFS(gr, i, 1);

    cout<<"\nPass 1 completed\n";

    fill(all(v), 0);
    fill(all(childsVisited), 0);

    cout<<"\nInside pass 2\n";

    for(int i = N - 1; i >= 0; i--)
        if(v[ ft[i] ] != 1)
            DFS(g, ft[i], 2);

    cout<<"\nPass 2 completed\n";
}

int main()
{
    freopen("input.txt","r",stdin);
    vvi g(N), gr(N);
    //ifstream file("/home/tauseef/Desktop/DAA/SCC.txt");
    int first, second;
    //string line;
    unsigned long int cnt = 0;

    //while(getline(file,line,'\n')) //Reading from file
    //{
        //stringstream ss(line);
        //ss >> first;
        //ss >> second;
        //if(first == second) //Eliminating self loops
            //continue;
    for ( int i = 0; i < 5105043; ++i ){
        int first, second;
        scanf("%d %d",&first,&second);
        g[first-1].push_back(second-1); //Creating G & Grev
        gr[second-1].push_back(first-1);
    }
        //cnt++;
    //}

    cout<<"\nfile read successfully\n";


    kosaraju(g, gr);

    cout<<"\nFinishing order is: ";

    sort(size.rbegin(), size.rend()); //Sorting the SCC sizes in descending order

    cout<<"\nThe largest 5 SCCs are: ";

}