(简单) POJ 1797 Heavy Transportation，Dijkstra。

　　Description

Background
　　Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.

Problem
　　You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.

 

　　题目就是求最大路。。。

 

代码如下：

#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdio>

#define min(a,b) (a<b ? a:b)

using namespace std;

const int INF=10e8;
const int MaxN=1010;

struct Node
{
    int v,val;

    Node(int _v=0,int _val=0):v(_v),val(_val) {}
    bool operator < (const Node &a) const
    {
        return val<a.val;
    }
};

struct Edge
{
    int v,cost;

    Edge(int _v=0,int _cost=0):v(_v),cost(_cost) {}
};

vector <Edge> E[MaxN];

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

    for(int i=1;i<=n;++i)
    {
        lowcost[i]=0;
    }
    lowcost[start]=INF;

    que.push(Node(start,INF));

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

        u=qtemp.v;



        len=E[u].size();

        for(int i=0;i<len;++i)
        {
            v=E[u][i].v;
            cost=E[u][i].cost;

            if(min(cost,lowcost[u])>lowcost[v])
            {
                lowcost[v]=min(cost,lowcost[u]);
                que.push(Node(v,lowcost[v]));
            }
        }
    }
}

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

int ans[MaxN];

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

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

    scanf("%d",&T);

    for(int cas=1;cas<=T;++cas)
    {
        scanf("%d %d",&N,&M);

        for(int i=1;i<=N;++i)
            E[i].clear();

        for(int i=1;i<=M;++i)
        {
            scanf("%d %d %d",&a,&b,&c);

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

        Dijkstra(ans,N,1);

        printf("Scenario #%d:
",cas);
        printf("%d

",ans[N]);
    }

    return 0;
}