(简单) POJ 2387 Til the Cows Come Home，Dijkstra。

　　Description

　　 Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

　　Farmer John's field has N (2 <= N <= 1000)
landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the
apple tree grove in which Bessie stands all day is landmark N. Cows
travel in the field using T (1 <= T <= 2000) bidirectional
cow-trails of various lengths between the landmarks. Bessie is not
confident of her navigation ability, so she always stays on a trail from
its start to its end once she starts it.

　　Given the trails between the landmarks, determine the
minimum distance Bessie must walk to get back to the barn. It is
guaranteed that some such route exists.

 

　　最短路模板题。。。

 

代码如下：

#include<iostream>
#include<cstring>
#include<queue>

using namespace std;

/////////////////////////////////////////////////////////////////

const int MaxN=;
const int INF=10e8;

struct Node
{
    int v,val;

    Node(int _v=,int _val=):v(_v),val(_val) {}
    bool operator < (const Node &a) const
    {
        return val>a.val;
    }
};

struct Edge
{
    int v,cost;

    Edge(int _v=,int _cost=):v(_v),cost(_cost) {}
};

vector <Edge> E[MaxN];
bool vis[MaxN];

void Dijkstra(int lowcost[],int n,int start)
{
    priority_queue <Node> que;
    Node qtemp;
    int len;
    int u,v,cost;

    for(int i=;i<=n;++i)
    {
        lowcost[i]=INF;
        vis[i]=;
    }
    lowcost[start]=;

    que.push(Node(start,));

    while(!que.empty())
    {
        qtemp=que.top();
        que.pop();

        u=qtemp.v;

        if(vis[u])
            continue;

        vis[u]=;

        len=E[u].size();

        for(int i=;i<len;++i)
        {
            v=E[u][i].v;
            cost=E[u][i].cost;

            if(!vis[v] && lowcost[v]>lowcost[u]+cost)
            {
                lowcost[v]=lowcost[u]+cost;
                que.push(Node(v,lowcost[v]));
            }
        }
    }
}

inline void addEdge(int u,int v,int c)
{
    E[u].push_back(Edge(v,c));
}

/////////////////////////////////////////////////////////////////

int ans[];

int main()
{
    ios::sync_with_stdio(false);

    int N,T;
    int a,b,c;

    cin>>T>>N;

    for(int i=;i<=T;++i)
    {
        cin>>a>>b>>c;

        addEdge(a,b,c);
        addEdge(b,a,c);
    }

    Dijkstra(ans,N,N);

    cout<<ans[]<<endl;

    return ;
}