POJ 3225 Roadblocks

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贴一道利用优先队列实现的Dijkstra算法。这道题是要求次短路，所以对Dijkstra算法略作修改，同时保存最短和次短路数组。

#include<cstdio>#include<iostream>#include<vector>#include<queue>#include<functional>//#include<xutility>using namespace std; struct Edge{    long long cost;    int to;    Edge(long long co, int t):cost(co),to(t){}};typedef pair< long long, int >  P;int main(){    int r,n;    const long long INF=1000000000;    scanf("%d%d",&n,&r);    vector<Edge> *edge=new vector<Edge>[n+1];    long long *d=new long long [n+1];    long long *d1=new long long [n+1];    for(int i=0;i<r;i++)    {        int a,b;        long long cost;        scanf("%d%d%lld",&a,&b,&cost);        edge[a].push_back(Edge(cost,b));        edge[b].push_back(Edge(cost,a));    }    priority_queue< P, vector<P>, greater<P> > que;     fill(d+1,d+n+1,INF);    fill(d1+1,d1+n+1,INF);    d[1]=0;     que.push(make_pair(0,1));    while(!que.empty())    {        P p=que.top();        que.pop();        int v=p.second;        if(d1[v]<p.first)            continue;        for( int i=0;i<edge[v].size();i++)        {            Edge e = edge[v][i];            long long dTmp=p.first+e.cost;            if(d[e.to]>dTmp){                swap(d[e.to],dTmp);                que.push(make_pair(d[e.to],e.to));            }            if(d1[e.to]>dTmp && d[e.to]<dTmp)            {                d1[e.to]=dTmp;                que.push(make_pair(d1[e.to],e.to));            }        }    }     cout<<d1[n]<<endl;     return 0;}

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