POJ 3255 Roadblocks

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链接：http://poj.org/problem?id=3255

题目要求1~N的次短路径.

代码来自《挑战程序设计竞赛》

#include <iostream>#include<queue>#include<vector>#define MAX_N 100005#define INF 1000000using namespace std;struct edge{    int to;    int cost;};typedef pair<int ,int> P;vector<edge> g[MAX_N];int dist1 [MAX_N];int dist2 [MAX_N];priority_queue<P,vector<P>,greater<P> > que;void solve(int N){    int i;    for(i=0;i<MAX_N;i++)    {        dist1[i]=INF;        dist2[i]=INF;    }    dist1[0]=0;    que.push(make_pair(0,1));    while(!que.empty())    {        P p=que.top();        que.pop();        int v=p.second,d=p.first;        if(dist2[v]<d)            continue;        for(i=0; i<g[v].size(); i++)        {            edge &e=g[v][i];            int d2=d+e.cost;            if(dist1[e.to]>d2)            {                swap(dist1[e.to],d2);                que.push(make_pair(dist1[e.to],e.to));            }            if(dist2[e.to]>d2&&dist1[e.to]<d2)            {                dist2[e.to]=d2;                que.push(make_pair(dist2[e.to],e.to));            }        }    }    cout<<dist2[N]<<endl;}int main(){    int N,R;    int ta,tb,tc;    int i;    edge e;    cin>>N>>R;    for(i=0;i<R;i++)    {        cin>>ta>>tb>>tc;        e.to=tb;e.cost=tc;        g[ta].push_back(e);        e.to=ta;        g[tb].push_back(e);    }    solve(N);    return 0;}

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