hdu 2544 最短路

最短路的入门题

朴素的dij写了一遍之后拿优先队列又写了一遍

#include <iostream>#include <cstdio>#include <map>#include <queue>#include <vector>using namespace std;struct edge{    int to, dist;    edge(int v, int d) :        to(v), dist(d) {}};struct node{    int d, from;    node(int dd, int too):        d(dd), from(too) {}    bool operator < (const node& rhs) const{        return d > rhs.d;    }};const int maxn = 1e6;const int inf = 0x3f3f3f3f;vector <edge> edges[maxn];int d[maxn];int n;void dijkstra(int s){    priority_queue <node> q;    for (int i = 0; i != n; ++i)        d[i] = inf;    d[s] = 0;    q.push(node(0,s));    while (!q.empty())    {        node x = q.top();        q.pop();        if (x.d != d[x.from]) //若该点已出现过则跳过 类似与朴素的dij中的vis数组。            continue;        for (int i = 0; i != edges[x.from].size(); ++i)        {            edge& e = edges[x.from][i];  //对从from出发的各条边进行松弛操作            if (d[e.to] > d[x.from] + e.dist)            {                d[e.to] = d[x.from] + e.dist;                q.push(node(d[e.to], e.to));            }        }    }}int main(){    int k;    while (cin >> n  >> k && n)    {        int p1, p2, dis;        for (int i = 0; i != k; ++i)  //无向图 建立双向边        {            cin >> p1 >> p2 >> dis;            edges[p1-1].push_back(edge(p2-1,dis));            edges[p2-1].push_back(edge(p1-1,dis));        }        dijkstra(0);        if (d[n-1] == inf)            cout << "-1" << endl;        else            cout << d[n-1] << endl;        for (int i = 0; i != 2*k; ++i)            edges[i].clear();    }    return 0;}