POJ - 1797 Heavy Transportation (变形Dijkstra)

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本文介绍了一个货运路径优化问题，目标是从起点到终点找到允许通过的最大重量路径。利用一种改进的Dijkstra算法，确保所选路径中最小承载能力的边最大化，从而实现货物的最大运输量。

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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

Sample Output

Scenario #1:
4

题目大意：
把货物从城市1运往城市n，要使能运载的货物最多。有m条路，每一条路从城市a到城市b运货量为x

思路：要使所走路径通货量最小的一条路的通货量最大

dij变形 dis[j]=max(dis[j],min(dis[k],mp[k][j]));

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 1005
#define inf 0x3f3f3f
using namespace std;
int map[N][N],dis[N],vis[N];
int n;
int dij(int s,int e)
{
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    for(int i=1;i<=n;i++)
            dis[i]=map[s][i];
    vis[s]=1;
    int ans,u;
    for(int i=1;i<=n;i++)
    {
       ans=0;
        for (int j=1;j<=n;j++)
            if (!vis[j]&&dis[j]>ans)
            {
                ans=dis[j];
                u=j;
            }
        vis[u]=1;
        for (int j=1;j<=n;j++)
        {
            if(map[u][j]>-1)
            dis[j]=max(dis[j],min(dis[u],map[u][j]));
        }
    }
    return dis[e];
}
int main()
{
    int t,m,a,b,x,k=1;
    scanf("%d",&t);
    while (t--)
    {
        memset(map,-1,sizeof(map));
        scanf("%d %d",&n,&m);
        for (int i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&x);
            map[a][b]=map[b][a]=x;
        }
        printf("Scenario #%d:\n%d\n\n",k++,dij(1,n));
    }
    return 0;
}