poj1797 dijsktra变形

2024-04-28 17:18

文章标签 变形 poj1797 dijsktra

本文主要是介绍poj1797 dijsktra变形，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

如题：http://poj.org/problem?id=1797

Heavy Transportation

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 21783		Accepted: 5793

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

Sample Output

Scenario #1:
4

Source

TUD Programming Contest 2004, Darmstadt, Germany

题意：给出一堆双向边，权值是本条路可运输的最大量。求出最大能运输多重的东西

思路：不难发现，一条路径最大运输量取决于这条路径的最小权值边。对dijsktra算法松弛操作进行修改。

dis[i]代表到当前顶点i的这条路径的最大承重值。

for(j=1;j<=n;j++)
{
if(vis[j]) continue;
dis[j]=MAX(dis[j],MIN(dis[k],a[k][j]));
}

松弛操作，理解一下就能懂。

#include<iostream>
using namespace std;
#define MAXN 1005
int a[MAXN][MAXN];
int dis[MAXN];
bool vis[MAXN];
int MAX(int a,int b)
{return a>b?a:b;}
int MIN(int a,int b)
{return a>b?b:a;}

void dij(int v,int n)
{
memset(vis,false,sizeof(vis));
int i,j;
for(i=1;i<=n;i++)
  dis[i]=a[v][i];
vis[v]=true;
for(i=1;i<n;i++)
{
  int temp=-1,k=v;
  for(j=1;j<=n;j++)
  {
   if(vis[j]) continue;
   if(temp<dis[j])
   {
    temp=dis[j];
    k=j;
   }
  }
  vis[k]=true;
  for(j=1;j<=n;j++)
  {
   if(vis[j]) continue;
   dis[j]=MAX(dis[j],MIN(dis[k],a[k][j]));
  }
}
}
int main()
{
// freopen("C:\\1.txt","r",stdin);
int N;
cin>>N;
int count=0;
while(N--)
{
  count++;
  memset(a,-1,sizeof(a));
  int n,m;
  scanf("%d %d",&n,&m);
  int i;
  for(i=1;i<=m;i++)
  {
   int t1,t2,w;
   scanf("%d %d %d",&t1,&t2,&w);
   a[t1][t2]=w;
   a[t2][t1]=w;
  }
  dij(1,n);
  printf("Scenario #%d:\n%d\n",count,dis[n]);
  printf("\n");
}
}