dijsktra求最短路徑
阿新 • • 發佈:2022-06-06
講演算法原理的有很多,直接貼程式碼
dijkstra演算法是直接對鄰接矩陣進行操作求出最短路徑的,我專案中的圖結構需要轉化成鄰接矩陣,所以會有下面程式碼
圖結構是一個map,first表示節點的index,second是一個結構體,包含節點的詳細資訊
map<int, node> routeTable;
struct node { public: string ip; //節點的ip int port; //節點的埠 vector<OP_TYPE> ops; //節點支援的OP floatoverload; //節點的負載 bool status; // 節點的狀態 int ncmIndex; //ncm index int reportCount; vector<pair<int,int>> adjacencyIndex; //鄰接點的index及距離 };
所以需要根據這個map來生成一個鄰接矩陣,dijkstra演算法是從圖的起點找出它與所有點的最短路徑,而我的需求是給定兩個點求他們在圖上的最短距離,所以生成鄰接矩陣不能按map本來的順序去生成,需要將輸入的兩個點作為鄰接矩陣的第一個點和最後一個點。舉例來說如果原來節點的順序是[0,1,2,3,4,5,6],我們想求2和4在圖上的最短距離,那麼就需要將map便利的順序變成[2,0,1,3,5,6,4]
生成鄰接矩陣
std::vector<std::vector<int>> GlobalRouteTable::getRouteAdjacentMatrix( int startNodeIndex, int endNodeIndex) { std::vector<int> graphNodeOrder = {}; // 將startNodeIndex和endNodeIndex對應的點作為圖遍歷的起點和重點 // 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; // 初始化鄰接矩陣 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); } // 對角線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; }
求最短路徑
int GlobalRouteTable::dijsktra(std::vector<std::vector<int>> &&graph_matrix) { // init // 目標點到其他點的距離 std::vector<int> dist = {}; // 是否找到最短路徑 std::vector<bool> target = {}; // 初始化圖的第一個節點並錄入與之相鄰的節點的距離 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]; // 確定下一個要搜尋的點 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]; }
函式呼叫
int GlobalRouteTable::getNodeDist(int startNodeIndex, int endNodeIndex) { if (startNodeIndex == endNodeIndex) return 0; return dijsktra(std::move(getRouteAdjacentMatrix(startNodeIndex, endNodeIndex))); }
圖結構
除了標記的變長是4,其餘的邊長都是1
測試用例
#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(); }