Dijkstra模板

lrj紫书十一章

设m edges,n vertexs

复杂度 mlog(n)

算法中以下部分使得每个边都被遍历到,m。而优先队列插入复杂度为log(n).故整体mlog(n)

注意m可能大于n^2 最后复杂的>n^2.但不常见

 while(!Q.empty()) //第一轮将s能到的点都压入队列,之后每次取d[i]最小的点先出(优先队列)
        {
            HeapNode x=Q.top();Q.pop();
            int u=x.u;                     //u 当前处理点编号
            if(done[u])continue;
            done[u]=true;
            for(int i=0;i<G[u].size();i++)
            {
                Edge &e=edges[G[u][i]];
                if(d[e.to]>d[u]+e.dist)
                {
                    d[e.to]=d[u]+e.dist; //d[u],出发点s到u的距离
                    p[e.to]=G[u][i];     //到达e.to点的边为G[u][i]
                    Q.push(HeapNode(d[e.to],e.to));
                }
            }
        }
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;
const int maxn = 300 + 5;
const int INF = 99999999;
struct Edge
{
    int from,to,dist;
    Edge(int f=0,int t=0,int d=0):from(f),to(t),dist(d){}
};
struct HeapNode//优先队列节点
{
    int d,u;
    HeapNode(int _d=0,int _u=0):d(_d),u(_u){}
    bool operator<(const HeapNode &rhs)const
    {
        return d>rhs.d;
    }
};
struct Dijkstra  //边权为正 负权存在用Bellman-Ford 每两点间最短路floyd
{
    int n,m;               //点数和边数  O(mlog n)
    vector<Edge> edges;    //边列表
    vector<int> G[maxn];   //每个节点出发的边编号(编号从0开始)
    bool done[maxn];       //是否已永久标号
    int d[maxn];           //s到各个点的距离
    int p[maxn];           //最短路中的上一条边
    void init(int n)
    {
        this->n=n;
        for(int i=0;i<n;i++)G[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,int dist)
    {
        edges.push_back(Edge(from,to,dist));
        m=edges.size();
        G[from].push_back(m-1);
    }

    void dijkstra(int s)   //s start
    {
        priority_queue<HeapNode> Q;   //优先队列,d[i]越小越先出队
        for(int i=0;i<n;i++)d[i]=INF;
        d[s]=0;
        memset(done,0,sizeof(done));
        Q.push(HeapNode(0,s));
        while(!Q.empty()) //第一轮将s能到的点都压入队列,之后每次取d[i]最小的点先出(优先队列)
        {
            HeapNode x=Q.top();Q.pop();
            int u=x.u;                     //u 当前处理点编号
            if(done[u])continue;
            done[u]=true;
            for(int i=0;i<G[u].size();i++)
            {
                Edge &e=edges[G[u][i]];
                if(d[e.to]>d[u]+e.dist)
                {
                    d[e.to]=d[u]+e.dist; //d[u],出发点s到u的距离
                    p[e.to]=G[u][i];     //到达e.to点的边为G[u][i]
                    Q.push(HeapNode(d[e.to],e.to));
                }
            }
        }
    }
};

int main()
{
    Dijkstra dijk;
    int n,m;                 //n number of vertex m number of edges
    while(scanf("%d%d", &n, &m)==2) {
        dijk.init(n);
        int from,to,dist;
        for(int i = 0; i < m; i++) {
            scanf("%d%d%d",&from,&to,&dist);
            dijk.AddEdge(from,to,dist);
        }
    dijk.dijkstra(0);
    for(int i=0;i<n;i++)
        cout<<"from 0 to "<<i<<"'s minimum distance is :"<<dijk.d[i]<<endl;
    }
    return 0;
}

posted @ 2019-10-16 21:11  Erio  阅读(210)  评论(0编辑  收藏  举报