使用优先队列（Priority Queue）：通过使用优先队列来存储未访问顶点的集合，我们可以将算法的时间复杂度降...
I was trying to solveCSES Shortest Routes Iusing priority_queue. However, I faced TLE even though I was storing negative of distance in it. After a bit of reading onCP-Algo's Page, they said that The main difference to the implementation with set is that in many languages, including C++...
使用priority_queue实现Dijkstra 主要是记得传比较算子进去,优先队列要的那个算子参数大概是用来定义优先级的,默认是less,大概是优先级低的先出队。想要从小到大排序要传greater进去。 1priority_queue< edge,vector<edge>,greater<edge> >Q;2Q.push(make_pair(0,1));dis[1]=0;3while(!Q.empty()){4intu=...
AI代码解释 #include<iostream>#include<vector>#include<queue>using namespace std;constintINF=0x3f3f3f3f;// 定义正无穷// 定义图的邻接表表示typedef pair<int,int>P;// first表示节点编号,second表示边权值vector<vector<P>>graph;// 迪杰斯特拉算法函数voiddijkstra(int start,vector<int>&dist,vector<i...
Dijkstra-priority_queue #include <cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<vector>#include#include<set>#include<string>usingnamespacestd;//---#defineCL(a,num) memset(a,num,sizeof(a));#defineBtoA(x,y) memcpy(x,y,sizeof(x));#defineeps 1e-12#...
} 1. 2. 3. 4. 5. 6. 7. 8. 而优先队列在这里做的一点就是,能够在查找最小未访问的点时,能够快速找到,使得 第一步能够以 logn的复杂度找到最小未访问点u #include<iostream> #include<vector> #include<cstring> #include<queue> using namespace std; ...
using namespace std; //矩阵,存储图 int graph[100][100]; //顶点、边数 int v,e; //优先队列,使用数组 int pri[100]; //存储起点到其它顶点之间的最短距离 int dis[100]; //设置无穷大常量 int const INF =INT_MAX; /* *初始化函数 ...
{ return w>x.w;//这里注意符号要为'>' }};priority_queue<node>q;//优先队列,其实这里一般使用一个pair,但为了方便理解所以用的结构体inline void add(int u,int v,int w){ e[++cnt].u=u; //这句话对于此题不需要,但在缩点之类的问题还是有用的 e[cnt].v=v; e[cnt].w=w; e[cnt]....
remark:朴素dijkstra算法用于求稠密数据的最短路。 #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int N = 510; int n, m; int g[N][N]; int dist[N]; bool st[N]; int dijkstra() { memset(dist, 0x3f, sizeof dist); dist[1] = 0; for (int...
#include <bits/stdc++.h> using namespace std; const int INF = 0x7ffffff; int vertexNum, edgeNum; vector<vector<int> > d(105, vector<int>(105, INF)); int Floyed() { for (int k = 1; k <= vertexNum; ++k) { for (int i = 1; i <= vertexNum; ++i) { for (int j =...