Heavy Transportation

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
題意：有n個城市，m條道路，在每條道路上有一個承載量，現在要求從1到n城市最大承載量，而最大承載量就是從城市1到城市n所有通路上的最大承載量（每條通路的承載量是該通路上所有道路的承載量的最小值，例：1->2承載量為3，2->3承載量為1，則1->3的承載量為1）
Dijkstra是求最短路的演算法，這裡變形一下（因為要求最大承載量，要使每條邊的權值儘量大）

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1005;
const int Inf=999999999;
int n,m;
int G[maxn][maxn],dis[maxn];
bool vis[maxn];
void dj(){
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	dis[i]=G[1][i];
	vis[1]=true;
	for(int k=
 
1;k<n;k++){
		int maxt=-1,u;
	    for(int i=1;i<=n;i++){
	    	if(!vis[i]&&dis[i]>maxt){//這裡是大於號
	    		maxt=dis[i];
	    		u=i;
			}
		}
		vis[u]=true;
	    for(int i=1;i<=n;i++){
	    	if(dis[i]<min(maxt,G[u][i]))//注意鬆弛條件
	    	dis[i]=min(maxt,G[u][i]);
		}
	}
	printf("%d\n\n",dis[n]);	
}
int main(){
	int T,u,v,w;
	scanf("%d",&T);
	for(int k=1;k<=T;k++){
		scanf("%d%d",&n,&m);
		memset(G,0,sizeof(G));
		for(int i=1;i<=m;i++){
			scanf("%d%d%d",&u,&v,&w);
			G[u][v]=G[v][u]=w;//雙向聯通
		}
		printf("Scenario #%d:\n",k);
		dj();
	}
	return 0;
}