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
//注意题上最后让输出2个换行
//#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;
const int INF=1e9+7;
const int maxn=1100;
typedef long long ll;
int t,n,m;
int vis[maxn];
int par[maxn];
int tu[maxn][maxn];
int dijkstra()
{
int i,j,k;
memset(vis,0,sizeof(vis));
for(i=1; i<=n; i++)
par[i]=tu[1][i];
for(i=1; i<=n; i++)
{
int minn=-1;
for(j=1; j<=n; j++)
{
if(!vis[j]&&minn<par[j])
{
k=j;
minn=par[j];
}
}
vis[k]=1;
for(j=1; j<=n; j++)
{
par[j]=max(par[j],min(tu[k][j],par[k]));
// if(!vis[j]&&par[j]<min(tu[k][j],par[k]))
// {
// par[j]=min(tu[k][j],par[k]);
// }
}
}
return par[n];
}
int main()
{
int cot=1;
scanf("%d",&t);
while(t--)
{
int i,j,a,b,c;
scanf("%d %d",&n,&m);
memset(tu,0,sizeof(tu));
while(m--)
{
scanf("%d %d %d",&a,&b,&c);
tu[a][b]=tu[b][a]=c;
}
printf("Scenario #%d:
",cot++);
printf("%d
",dijkstra());
}
return 0;
}