pku 2253は初期から終点までの最小の経路(一部)を記録し、ディジェストラは最短ルートを求める

  
  
  
  

   
   
   
   
 
    C - Heavy Transportation 
   
   
   
   
   Time Limit:3000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u
   
   
   
   
 
    Submit  
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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
 
    
 
     
1
3 3
1 2 3
1 3 4
2 3 5
  
Sample Output 
Scenario #1:
4
  
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define inf 0xffffff
#define MAX 0x7fffffff
#include<iostream>
using namespace std;
#define MIN -0x7fffffff
__int64 mp[1222][1222];
__int64 dis[1222];
int visit[1222];

int n,m;
void suan()
{
    memset(visit,0,sizeof(visit)); visit[1]=1;
    for(int i=2; i<=n; ++i)
        dis[i]=MIN;
    dis[1]=MAX;


    int pos=1;
    for(int i=1; i<=n; i++)
    {
int mincost=0;
        for(int z=1; z<=n; ++z)
        {
            if(!visit[z]&&mincost<dis[z])
            {
                mincost=dis[z];
                pos=z;
            }
        }
        visit[pos]=1;

        for(int  j=1; j<=n; j++)
        {
            if(mp[pos][j]!=MAX)
            {
                dis[j]=max(dis[j],min(dis[pos],mp[pos][j]));

            }
        }

    }

}
int   main()
{
    int  test;

    scanf("%d",&test);
    for(int ii=1; ii<=test; ii++)
    {
        scanf("%d%d",&n,&m);
        for(int  i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                 mp[i][j]=mp[j][i]=MAX;
            }
        }
        int a,b,c;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);

            mp[a][b]=mp[b][a]=c;

        }
        suan();
        printf("Scenario #%d:
",ii);
        printf("%I64d

",dis[n]);
    }

}