Heavy Transportation

题目大意是就是何处一个图，n个顶点和m条边，每个边都有最大承载量，现在我要从1点运送货物到n点，求能运送货物的最大重量。

对于数据，第一行为t代表测试数据个数，第二行为n和m（意义见上），接着m行，每行三个整数分别是代表一条边的起点，终点及最大承重量。输出能运送货物的最大重量，格式见样例。注意数据输完后还要再多输一个空行。

对于数据，从1运到3有两种方案

方案1：1-2-3，其中1-2承重为3，2-3承重为5，则可以运送货物的最大重量是3（当大于3时明显1到不了2）

方案2：1-3，可知1-3承重为4，故此路可运送货物的最大重量是4，故答案输出4

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

13 31 2 31 3 42 3 5

Sample Output

Scenario #1:
4
#include <iostream>#include<cstdio>using namespace std;#define MAXV 1010#define min(a,b) (a<b?a:b)int map[MAXV][MAXV],n,m;int dijkstra(){int vis[MAXV],d[MAXV],i,j,v;for(i=1;i<=n;i++){vis[i]=0;d[i]=map[1][i];}for(i=1;i<=n;i++){int f=-1;for(j=1;j<=n;j++)if(!vis[j] && d[j]>f){f=d[j];v=j;}vis[v]=1;for(j=1;j<=n;j++)if(!vis[j] && d[j]<min(d[v],map[v][j])){d[j]=min(d[v],map[v][j]);}}return d[n];}int main(){int t,i,j,sum,a,b,c;scanf("%d",&sum);for(t=1;t<=sum;t++){scanf("%d%d",&n,&m);for(i=0;i<=n;i++)for(j=0;j<=n;j++)map[i][j]=0;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);map[a][b]=map[b][a]=c;}printf("Scenario #%d:\n",t);printf("%d\n\n",dijkstra());}return 0;}