8.6 E - Heavy Transportation

                                 E - 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

题意：一个城市有n个点，这n个点之间有m条路，每条路都有一个最大承重量，现在需要从第一个点出发把货物运送到第n个点，找到一条路，使运的货物最多。

样例解释：由样例可知，从1到3有2条路径，分别是1→2→3和1→3，这两天条路径所能通行的最大承重量分别为3 ，5，要使运输的货物最多，就需要寻找到的路径中桥梁的承重量的最小值是最大的，所以最大承载量为4。

解题思路：借助Dijkstra算法的思想，先找出与第一个点连通的其他点之间桥梁的最大承重，然后寻找找到的该点与其他点连通时两点之间桥梁的最下值更新数组。

#include<stdio.h>
#include<string.h>
int a[1100][1100],book[1100],d[1100];

int Min(int x,int y)
{
	if(x>y)
	{
		x=y;
	}
			
	return x;
}

int main()
{
	int i,j,k,m,n,c,max=-100000,min,t,T,f=1;
	int t1,t2,t3;
	while(scanf("%d",&T)!=EOF)
	{
		while(T--)
		{
			memset(a,0,sizeof(a));
			memset(d,0,sizeof(d));
			memset(book,0,sizeof(book));
			
			scanf("%d%d",&n,&m);
			
			
			for(i=1;i<=n;i++)
			{
				for(j=1;j<=n;j++)
				{
					if(i==j)
					{
						a[i][j]=0;
					}
					else
					{
						a[i][j]=max;
					}
				}
			}		//初始化
			
			for(i=1;i<=m;i++)
			{
				scanf("%d%d%d",&t1,&t2,&t3);
				a[t1][t2]=t3;//正反储存
				a[t2][t1]=t3;
			}	
			printf("Scenario #%d:\n",f);f++;
			for(i=1;i<=n;i++)
			{
				d[i]=a[1][i];
			}
			c=0;
			book[1]=1;
			c++;
			while(c<n)
			{
				min=max;
				for(i=1;i<=n;i++)//找到当前一行的最大值
				{
					if(book[i]==0&&d[i]>min)
					{
						min=d[i];
						j=i;
					}
						
				}
				book[j]=1;
				c++;
				for(i=1;i<=n;i++)//从寻找到的点向下寻找，判断是否更新最下值
				{
					t=Min(a[j][i],d[j]);
					if(book[i]==0&&d[i]<a[j][i])
					{
						d[i]=t;
					}
				}	
			}
			printf("%d\n\n",d[n]);
			
		}	
	}
	return 0;
}