poj 1797 Heavy transportation --最短路变形

 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.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

Sample Input

1
3 3
1 2 3
1 3 4
2 3 5

Sample Output

Scenario #1:
4

题意：
　　卡车从 1 点行驶到 n 点，卡车自身有一定的重量，每条路也都有一个最大的承重量。计算在每条路的承重范围之内，从 1 点到 n 点，
　　 可以经过的卡车的最大的重量是多少。
　　输入顶点数，路径数 和 每条路径的最大承重
　　输出该城市可以经过的最大的卡车的重量。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define maxn 3010
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
struct Node{ int to,cost; };
int Vex,Edge;
vector<Node> graph[maxn];
long long dis[maxn];
void Dijkstra(int x)
{
    priority_queue<Pair,vector<Pair> ,less<Pair> > que;
    fill(dis,dis+Vex+2,0);
    dis[x] = INF;
    que.push(Pair(INF,x));
    while(!que.empty()){
        Pair p = que.top();
        que.pop();
        int v = p.second;
        if(v == Vex) break;
        if(dis[v] > p.first) continue;
        for(int i = 0;i < graph[v].size();i++){
            Node node = graph[v][i];
            if(dis[node.to] < min(node.cost*1LL,dis[v])){
                dis[node.to] = min(node.cost*1LL,dis[v]);
                que.push(Pair(dis[node.to],node.to));
            }
        }
    }
}

int main(){
    int t;
    scanf("%d",&t);
    for(int test = 1;test<=t;test++){
        for(int i = 0;i < Vex;i++)graph[i].clear();
        int from,to,value;
        Node tmp_node;
        scanf("%d%d",&Vex,&Edge);
        for(int i = 0;i < Edge;i++){
            scanf("%d%d%d",&from,&to,&value);
            tmp_node.to = to;
            tmp_node.cost = value;
            graph[from].push_back(tmp_node);
            tmp_node.to = from;
            graph[to].push_back(tmp_node);
        }
        Dijkstra(1);
        printf("Scenario #%d:\n%lld\n",test,dis[Vex]);
        if(test != t)printf("\n");
    }
    return 0;
}