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dijsktra求最短路径

时间:2022-10-04 16:33:10浏览次数:47  
标签:getNodeDist matrix 路径 ASSERT dijsktra EQ ptr size

讲算法原理的有很多,直接贴代码

dijkstra算法是直接对邻接矩阵进行操作求出最短路径的,我项目中的图结构需要转化成邻接矩阵,所以会有下面代码

图结构是一个map,first表示节点的index,second是一个结构体,包含节点的详细信息

map<int, node> routeTable;
struct node
{
public:
string ip; //节点的ip
int port; //节点的端口
vector<OP_TYPE> ops; //节点支持的OP
float overload; //节点的负载
bool status; // 节点的状态
int ncmIndex; //ncm index
int reportCount;
vector<pair<int,int>> adjacencyIndex; //邻接点的index及距离

所以需要根据这个map来生成一个邻接矩阵,dijkstra算法是从图的起点找出它与所有点的最短路径,而我的需求是给定两个点求他们在图上的最短距离,所以生成邻接矩阵不能按map本来的顺序去生成,需要将输入的两个点作为邻接矩阵的第一个点和最后一个点。举例来说如果原来节点的顺序是[0,1,2,3,4,5,6],我们想求2和4在图上的最短距离,那么就需要将map便利的顺序变成[2,0,1,3,5,6,4]

生成邻接矩阵

std::vector<std::vector<int>> GlobalRouteTable::getRouteAdjacentMatrix(
int startNodeIndex, int endNodeIndex)
{
std::vector<int> graphNodeOrder = {};
// 将startNodeIndex和endNodeIndex对应的点作为图遍历的起点和重点
// if(routeTable.find(startNodeIndex)==routeTable.end()||routeTable.find(endNodeIndex)==routeTable.end())
graphNodeOrder.push_back(startNodeIndex);
for (auto it : routeTable) {
if (it.first != startNodeIndex && it.first != endNodeIndex
&& it.second.adjacencyIndex[0].first != -1)
graphNodeOrder.push_back(it.first);
}
graphNodeOrder.push_back(endNodeIndex);
std::vector<std::vector<int>> adjacent_matrix;
// 初始化邻接矩阵
for (size_t i = 0; i < graphNodeOrder.size(); ++i) {
std::vector<int> matrix_line = {};
for (size_t j = 0; j < graphNodeOrder.size(); ++j)
matrix_line.push_back(65535);
adjacent_matrix.push_back(matrix_line);
}
// 对角线0
for (size_t i = 0; i < graphNodeOrder.size(); ++i)
adjacent_matrix[i][i] = 0;
for (size_t i = 0; i < graphNodeOrder.size(); ++i) {
for (auto j : routeTable[graphNodeOrder[i]].adjacencyIndex) {
int tmp_index = 0;
for (auto k = graphNodeOrder.begin(); k != graphNodeOrder.end();
++k) {
if (*k != j.first)
tmp_index += 1;
else
break;
}
adjacent_matrix[i][tmp_index] = j.second;
}
}
return adjacent_matrix;
}

求最短路径

int GlobalRouteTable::dijsktra(std::vector<std::vector<int>> &&graph_matrix)
{
// init
// 目标点到其他点的距离
std::vector<int> dist = {};
// 是否找到最短路径
std::vector<bool> target = {};
// 初始化图的第一个节点并录入与之相邻的节点的距离
for (size_t i = 0; i < graph_matrix.size(); ++i) {
dist.push_back(graph_matrix[0][i]);
target.push_back(false);
}
target[0] = true;
int min, k;
// start search
for (size_t i = 1; i < graph_matrix.size(); ++i) {
min = 65535;
for (size_t j = 1; j < graph_matrix.size(); ++j) {
if (!target[j] && min > dist[j]) {
min = dist[j];
// 确定下一个要搜索的点
k = j;
}
}
target[k] = true;
for (size_t j = 1; j < graph_matrix.size(); ++j) {
if (!target[j] && dist[j] > dist[k] + graph_matrix[k][j]) {
dist[j] = dist[k] + graph_matrix[k][j];
}
}
}
return dist[dist.size() - 1];
}

函数调用

int GlobalRouteTable::getNodeDist(int startNodeIndex, int endNodeIndex)
{
if (startNodeIndex == endNodeIndex) return 0;
return dijsktra(std::move(getRouteAdjacentMatrix(startNodeIndex, endNodeIndex)));
}

图结构

除了标记的变长是4,其余的边长都是1

dijsktra求最短路径_最短路径

 

 测试用例

#include "../../src/routeTable.h"

#include <gtest/gtest.h>

TEST(routetableTest, checkGetDist)
{
auto ptr = GlobalRouteTable::GetInstance();
std::cout << "load table result: " << ptr->loadRouteTable() << std::endl;
//
ASSERT_EQ(ptr->getNodeDist(0, 0), 0);
ASSERT_EQ(ptr->getNodeDist(0, 1), 1);
ASSERT_EQ(ptr->getNodeDist(0, 2), 1);
ASSERT_EQ(ptr->getNodeDist(2, 1), 2);
ASSERT_EQ(ptr->getNodeDist(2, 3), 1);
ASSERT_EQ(ptr->getNodeDist(2, 6), 3);
ASSERT_EQ(ptr->getNodeDist(0, 10), 4);
ASSERT_EQ(ptr->getNodeDist(0, 11), 5);
ASSERT_EQ(ptr->getNodeDist(1, 11), 6);
ASSERT_EQ(ptr->getNodeDist(0, 12), 5);
ASSERT_EQ(ptr->getNodeDist(0, 16), 6);
ASSERT_EQ(ptr->getNodeDist(2, 12), 6);
ASSERT_EQ(ptr->getNodeDist(2, 16), 7);
ASSERT_EQ(ptr->getNodeDist(6, 16), 8);
}
int main(int argc, char **argv)
{
::testing::InitGoogleTest(&argc, argv);
return RUN_ALL_TESTS();
}

dijsktra求最短路径_邻接矩阵_02

 

无情的摸鱼机器



标签:getNodeDist,matrix,路径,ASSERT,dijsktra,EQ,ptr,size
From: https://blog.51cto.com/u_12870633/5731344

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