[std :: bad_alloc在大数据集的dijkstra计算中

问题描述 投票:0回答:1

我正在尝试使用dijkstra算法求解大图的最短路径。问题是,当我在CLion中执行程序时,我总是在节点491上获取std :: bad alloc,但是当我尝试在Ubuntu VM上执行相同操作时,我开始将内核转储。

我是C ++的新手,所以我很难理解为什么会发生这种情况。

这是我的代码:

实用程序:

#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <sstream>
#include <ctime>

#define INFINITY 9999999

int maxNode = 0;
using namespace std;

vector<int> loadFile(const string &path) {
    vector<int> graph;
    ifstream file;
    file.open(path);
    if (!file.fail()) {
        string line;
        while (getline(file, line)) {
            stringstream ss(line);
            for (int i; ss >> i;) {
                if (i + 1 > maxNode)
                    maxNode = i + 1;
                graph.push_back(i);
                if (ss.peek() == ';')
                    ss.ignore();
            }
        }
        file.close();
    }
    return graph;
}

int **formatGraph(vector<int> inData) {
    int **graph = 0;
    int currentIndex = 0;
    int srcNode = inData[0];
    int dstNode = inData[1];
    int cost = inData[2];
    graph = new int *[maxNode];
    for (int i = 0; i < maxNode; i++) {
        graph[i] = new int[maxNode];
        for (int j = 0; j < maxNode; j++) {
            if (srcNode == i && dstNode == j) {
                graph[i][j] = cost;
                currentIndex++;
                srcNode = inData[currentIndex * 3];
                dstNode = inData[currentIndex * 3 + 1];
                cost = inData[currentIndex * 3 + 2];
                //printf("%d %d\n", i, j);
            } else
                graph[i][j] = 0;
        }
    }
    for (int i = 0; i < maxNode; i++) {
        for (int j = 0; j < maxNode; j++) {
            graph[j][i] = graph[i][j];
        }
    }
    return graph;
}

算法:

void dijkstra(int **G, int n, int startnode) {
    printf("%d\n", startnode);
    int **cost = new int *[maxNode];
    int distance[maxNode], pred[maxNode];
    int visited[maxNode], count, mindistance, nextnode, i, j;
    for (i = 0; i < n; i++) {
        cost[i] = new int[maxNode];
        for (j = 0; j < n; j++)
            cost[i][j] = 0;
    }
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
            if (G[i][j] == 0)
                cost[i][j] = INFINITY;
            else
                cost[i][j] = G[i][j];
    for (i = 0; i < n; i++) {
        distance[i] = cost[startnode][i];
        pred[i] = startnode;
        visited[i] = 0;
    }
    distance[startnode] = 0;
    visited[startnode] = 1;
    count = 1;
    while (count < n - 1) {
        mindistance = INFINITY;
        for (i = 0; i < n; i++) {
            if (distance[i] < mindistance && !visited[i]) {
                mindistance = distance[i];
                nextnode = i;
            }
        }
        visited[nextnode] = 1;
        for (i = 0; i < n; i++) {
            if (!visited[i]) {
                if (mindistance + cost[nextnode][i] < distance[i]) {
                    distance[i] = mindistance + cost[nextnode][i];
                    pred[i] = nextnode;
                }
            }
        }
        count++;
    }
    delete[] cost;
    for (i = 0; i < n; i++)
        if (i != startnode) {
            j = i;
            do {
                j = pred[j];
            } while (j != startnode);
        }
}

这是我的主要功能:

int main() {
    vector<int> graph = loadFile("..\\data\\newFile2.csv");
    int **graphConverted = formatGraph(graph);
    //printMatrix(graphConverted);
    clock_t begin = clock();
    for (int i = 0; i < maxNode; i++)
        dijkstra(graphConverted, maxNode, i);
    clock_t end = clock();
    double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
    printf("\nTime: %f", elapsed_secs);
    return 0;
}

首先将数据加载到向量中,然后将其转换为邻接矩阵。数据以以下形式存储:

src_node; dst_node; cost1; 2; 31; 3; 301; 66; 20等

数据集由1004个节点和25571个边组成。

您能建议我任何解决方案来解决此问题吗?

c++ memory heap-memory dijkstra
1个回答
0
投票

dijkstra中,您在这里有动态内存分配:

int **cost = new int *[maxNode];

这里是i的循环:

cost[i] = new int[maxNode];

在此功能中,您只有一个呼叫delete[]

delete[] cost;

因此,保证了第二行new中的所有分配都将被泄漏。一段时间后,您将内存不足,导致显示std::bad_alloc

您需要将每个new[]呼叫与恰好是一个 delete[]呼叫相匹配。


完全不使用new / delete。而是将所有数组声明为std::vector,这将自动进行处理。

也不要使用可变长度数组,例如

int distance[maxNode], pred[maxNode];

它们是非标准的编译器扩展。也使这些std::vector

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