最短路径(2)--poj1797(Dijkstra变形题)

Heavy Transportation

                                              Time Limit:3000MS    Memory Limit:30000KB    64bit
IO Format:%lld & %llu

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

        

       这道题题目大意是就是一个图，n个顶点和m条边，每个边都有最大承载量，现在我要从1点运送货物到n点，求能运送货物的最大重量。(从1到n的某一条路径能运送货物的最大重量是该路径上每一条边中承重量最小的值，即最小边权，那么该题就是求所有路径中的最大的最小边权)

对于数据，第一行为t代表测试数据个数，第二行为n和m（意义见上），接着m行，每行三个整数分别是代表一条边的起点，终点及最大承重量。输出能运送货物的最大重量，格式见样例。注意数据输完后还要再多输一个空行。

       对于数据，从1运到3有两种方案:

方案1：1-2-3，其中1-2承重为3，2-3承重为5，则可以运送货物的最大重量是3（当大于3时明显1到不了2）

方案2：1-3，可知1-3承重为4，故此路可运送货物的最大重量是4，故答案输出4

        解题思路：其实这个求最大边可以近似于求最短路，只要修改下找最短路更新的条件就可以了。具体看代码：

最坑的是 我开始用的cin cout输入输出，超时了......

#include<stdio.h>
#include<string.h>
int N,M;
int mpt[1005][1005];
int dis[1005];
int vis[1005];
int Min(int a,int b)
{
return a>b?b:a;
}
void Dij()              //核心算法
{
int i,j;
for(i=0;i<=N-1;i++)
{
int max=-1;
int u=0;
for(j=1;j<=N;j++)
{
if(!vis[j]&&max<dis[j])
{
max=dis[j];
u=j;
}
}
if(u==0)break;
vis[u]=1;
for(j=1;j<=N;j++)
{
if(!vis[j]&&Min(mpt[u][j],dis[u])>dis[j])     //更新条件的变化
{
dis[j]=Min(dis[u],mpt[u][j]);
}
}
}
}
void Init()                //初始化
{
int i,j;
memset(vis,0,sizeof(vis));
for(i=1;i<=N;i++)
{
for(j=1;j<=N;j++)
{
mpt[i][j]=0;
}
}
}
int main()
{
int t;
scanf("%d",&t);
int cnt=0;
while(t--)
{
cnt++;
scanf("%d %d",&N,&M);
int i,j;
Init();
for(i=0;i<M;i++)
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
mpt[a][b]=mpt[b][a]=c;
}
for(i=1;i<=N;i++)
{
dis[i]=mpt[1][i];          //这个时候dis不代表从1到n的最短路径，而是最大承载量
}
Dij();
printf("Scenario #%d:\n",cnt);
printf("%d\n\n",dis
);
}
return 0;
}