如何为 2 个节点之间的单个最短路径优化 Dijkstra 算法? [英] How to optimize Dijkstra algorithm for a single shortest path between 2 nodes?

问题描述

我试图在 C 语言中理解 这个实现Dijkstra 算法，同时对其进行修改，以便仅找到 2 个特定节点(源和目标)之间的最短路径.

I was trying to understand this implementation in C of the Dijkstra algorithm and at the same time modify it so that only the shortest path between 2 specific nodes (source and destination) is found.

但是，我不知道到底要做什么.在我看来，没什么可做的，我似乎无法更改 d[] 或 prev[] 导致这些数组以最短的时间聚合了一些重要数据路径计算.

However, I don't know exactly what do to. The way I see it, there's nothing much to do, I can't seem to change d[] or prev[] cause those arrays aggregate some important data for the shortest path calculation.

我唯一能想到的就是在找到路径时停止算法，即在mini = destination标记为已访问时打破循环.

The only thing I can think of is stopping the algorithm when the path is found, that is, break the cycle when mini = destination when it's being marked as visited.

我还能做些什么来让它变得更好，或者就足够了吗?

Is there anything else I could do to make it better or is that enough?

虽然我很欣赏给我的建议，但人们仍然无法准确回答我提出的问题.我只想知道如何优化算法以仅搜索2 个节点之间的最短路径.目前，我对所有其他一般优化不感兴趣.我要说的是，在查找从节点 X 到所有其他节点的所有最短路径的算法中，如何优化它以仅搜索特定路径?

While I appreciate the suggestions given to me, people still fail to answer exactly what I questioned. All I want to know is how to optimize the algorithm to only search the shortest path between 2 nodes. I'm not interested, for now, in all other general optimizations. What I'm saying is, in an algorithm that finds all shortest paths from a node X to all other nodes, how do I optimize it to only search for a specific path?

PS:我刚刚注意到 for 循环从 1 开始，直到 <=，为什么不能从 开始>0 直到 <?

P.S: I just noticed that the for loops start at 1 until <=, why can't it start at 0 and go until <?

推荐答案

您问题中的实现使用了相邻矩阵，这导致 O(n^2) 实现.考虑到现实世界中的图通常是稀疏的，即节点数 n 通常非常大，但是边数远少于 n^2.

The implementation in your question uses a adjacent matrix, which leads O(n^2) implementation. Considering that the graphs in the real world are usually sparse, i.e. the number of nodes n is usually very big, however, the number of edges is far less from n^2.

您最好查看基于堆的 dijkstra 实现.

顺便说一句，单对最短路径的求解速度不能比来自特定节点的最短路径更快.

BTW, single pair shortest path cannot be solved faster than shortest path from a specific node.

#include<algorithm>
using namespace std;

#define MAXN 100
#define HEAP_SIZE 100
typedef int Graph[MAXN][MAXN];

template <class COST_TYPE>
class Heap
{
public:
    int data[HEAP_SIZE],index[HEAP_SIZE],size;
    COST_TYPE cost[HEAP_SIZE];
    void shift_up(int i)
    {
        int j;
        while(i>0)
        {
            j=(i-1)/2;
            if(cost[data[i]]<cost[data[j]])
            {
                swap(index[data[i]],index[data[j]]);
                swap(data[i],data[j]);
                i=j;
            }
            else break;
        }
    }
    void shift_down(int i)
    {
        int j,k;
        while(2*i+1<size)
        {
            j=2*i+1;
            k=j+1;
            if(k<size&&cost[data[k]]<cost[data[j]]&&cost[data[k]]<cost[data[i]])
            {
                swap(index[data[k]],index[data[i]]);
                swap(data[k],data[i]);
                i=k;
            }
            else if(cost[data[j]]<cost[data[i]])
            {
                swap(index[data[j]],index[data[i]]);
                swap(data[j],data[i]);
                i=j;
            }
            else break;
        }
    }
    void init()
    {
        size=0;
        memset(index,-1,sizeof(index));
        memset(cost,-1,sizeof(cost));
    }
    bool empty()
    {
        return(size==0);
    }
    int pop()
    {
        int res=data[0];
        data[0]=data[size-1];
        index[data[0]]=0;
        size--;
        shift_down(0);
        return res;
    }
    int top()
    {
        return data[0];
    }
    void push(int x,COST_TYPE c)
    {
        if(index[x]==-1)
        {
            cost[x]=c;
            data[size]=x;
            index[x]=size;
            size++;
            shift_up(index[x]);
        }
        else
        {
            if(c<cost[x])
            {
                cost[x]=c;
                shift_up(index[x]);
                shift_down(index[x]);
            }
        }
    }
};



int Dijkstra(Graph G,int n,int s,int t)
{
    Heap<int> heap;
    heap.init();
    heap.push(s,0);
    while(!heap.empty())
    {
        int u=heap.pop();
        if(u==t)
            return heap.cost[t];
        for(int i=0;i<n;i++)
            if(G[u][i]>=0)
                heap.push(i,heap.cost[u]+G[u][i]);
    }
    return -1;
}

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