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：有n个城市，m条道路，在每条道路上只有一个承载量，要求从1到n的最大承载量，最大承载量就是所有通路的最大承载量
比如：1->2 3； 2->3 5 ，那么1->3最大承载量为3
思路：首先找到和1相连的的点u和载重量最大值maxn，然后找出manx 和与u相连的点i之间的重量的最小值ans，最后拿这个最小值与dis[i]相比，如果ans>dis[i]，则需要更新，反复进行这个过程，知道外层循环结束
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m;
#define Inf 0x3f3f3f3f
const int N=1005;
int g[N][N];
bool vis[N];
int dis[N];
int Dijkstra()
{
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
dis[i]=g[1][i];
vis[1]=1;
for(int k=1;k<n;k++)
{
int maxn=0,u;
for(int i=1;i<=n;i++)
{
if(!vis[i]&&maxn<dis[i])
{
maxn=dis[i];
u=i;
}
}
vis[u]=1;
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
int ans=min(maxn,g[u][i]);
if(ans>dis[i])
dis[i]=ans;
}
}
}
return dis[n];
}
int main()
{
int t;
cin>>t;
for(int i=1;i<=t;i++)
{
cin>>n>>m;
memset(g,0,sizeof(g));
for(int j=0;j<m;j++)
{
int a,b,w;
cin>>a>>b>>w;
g[a][b]=g[b][a]=w;
}
printf("Scenario #%d:\n",i);
printf("%d\n\n",Dijkstra());
}
return 0;
}