openjudge heavy transportation

799:Heavy Transportation

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总时间限制: 

3000ms 

内存限制: 

65536kB

描述

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.

输入

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.

输出

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.

样例输入

1
3 3
1 2 3
1 3 4
2 3 5

样例输出

Scenario #1:
4

来源

TUD Programming Contest 2004, Darmstadt, Germany

题目大意：有N个点，M条路，每条路有一个最大载重量，要通过这条路，载重量要不超过该路的最大载重量。某某某要从点1到点N，求所有路径中载重量最大的那条路径。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,f[1100],t,num,road,ans[21000],ans1,vis[1100];
struct data
{
int x,y,len;
};
data street[1000003];
bool pd;
int i,j,k,next[21000],point[1100],u[21000],v[21000],w[21000];
int cmp(data a,data b)
{
return a.len>b.len;
}
int find(int x)
{
if (f[x]==x)  return x;
f[x]=find(f[x]);
return f[x];
}
void chuli(int a,int b,int c)
{
road++; u[road]=a; v[road]=b; w[road]=c; next[road]=point[a]; point[a]=road;
    road++; u[road]=b; v[road]=a; w[road]=c; next[road]=point[b]; point[b]=road;
}
void dfs(int x)
{
if (x==n)
{
  pd=true;
  return;
     }
int mp=point[x];
while (mp!=0)
{
 if (vis[v[mp]]==0)
  {
ans1++;
vis[v[mp]]=1;
ans[ans1]=w[mp];
dfs(v[mp]);
if (pd==true) return;
ans1--;
  }
 mp=next[mp];
     }
}
int cmp1(int a,int b)
{
return a<b;
}
int main()
{
scanf("%d",&t);
for (k=1;k<=t;k++)
{
scanf("%d%d",&n,&m);
for (i=1;i<=n;i++)
f[i]=i;
for (i=1;i<=m;i++)
scanf("%d%d%d",&street[i].x,&street[i].y,&street[i].len);
sort(street+1,street+m+1,cmp);
num=0; road=0;
memset(point,0,sizeof(point));  memset(next,0,sizeof(next));
memset(w,0,sizeof(w)); memset(u,0,sizeof(u)); memset(v,0,sizeof(v));
for (i=1;i<=m;i++)
{
if (find(street[i].x)!=find(street[i].y))
{
int r1=find(street[i].x); int r2=find(street[i].y);
f[r2]=r1;
num++;
chuli(street[i].x,street[i].y,street[i].len);
}
if (num==n-1)
 break;
}
memset(ans,0,sizeof(ans)); memset(vis,0,sizeof(vis));
pd=false; ans1=0;
vis[1]=1;
dfs(1);
sort(ans+1,ans+ans1+1,cmp1);
printf("Scenario #%d:\n",k);
printf("%d\n",ans[1]);
printf("\n");
}
}//这道题我采用了最小生成树的兄弟最大生成树，按同样的方法建树，只不过是每次选取最大的边，建树后从1到N的路径是唯一的，所以只需遍历一遍这条路径找出路径中的最小值即可。