dijkstra算法

//Dijkstra 算法的时间复杂度为O(n^2) 　　空间复杂度取决于存储方式，邻接矩阵为O(n^2)//非负权, 有向#include <iostream>#include <cstdio>using namespace std;const int max_vex = 100;const int max_dist = 100000;bool is_visited[max_vex];int adj[max_vex][max_vex];int  dist[max_vex];//节点i到节点1的最短距离, 0 作为一个标志位int vex_num;//节点数目void init(){memset(is_visited, 0x00, sizeof(is_visited));for (int i = 0; i < max_vex; ++i)dist[i] = max_dist;dist[1] = 0;// first set to zero, to start the process}int find_min_weight(){int min_vertex = 0;//first index is a flagfor (int i = 1; i <= vex_num; ++i){if (!is_visited[i] && dist[min_vertex] > dist[i])// 从未加入最小集合中，挑选距离最短的vertex加入最小集合{min_vertex = i;}}return min_vertex;}void dijkstra(){init();while(1){int p = find_min_weight();if (p == 0) break;is_visited[p] = true;// update flag to indicate p is added to the min setfor (int i = 1; i <= vex_num; ++i)// update the adjacent vertices{int dist_i = dist[p]+adj[p][i];if (!is_visited[i] && dist[i] > dist_i)dist[i] = dist_i;}}}int main(int argc, char* argv[]){for (int i = 0; i < max_vex; ++i){for (int j = 0; j < max_vex; ++j)adj[i][j] = max_dist;}adj[1][2] = 10;adj[1][5] = 5;adj[2][5] = 2;adj[2][3] = 1;adj[3][4] = 4;adj[4][3] = 6;adj[4][1] = 7;adj[5][4] = 2;adj[5][3] = 9;adj[5][2] = 3;vex_num = 5;dijkstra();for (i = 1; i <= 5; ++i)cout<<dist[i]<<" ";return 0;}