Heavy Transportation
| Time Limit: 3000MS | Memory Limit: 30000K | |
    |
|
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
Source
TUD Programming Contest 2004, Darmstadt, Germany
其实这个题看了好久都没搞懂意思，然后找了找博客原来就是给定n个站点（就这样理解吧），m条路，两个站点之间有且只有一条路，每条路都有一个最大载重量，求整个路径组成的图中的最大的承重；比如，1到2之间的载重是3，超过3便不能从这条路上过，而1到3之间的载重是4，这时已经将图连接起来了，所以最大值是4，如果选1->2->3,则1、2之间的路就不能过重量为5的升降机；
这样，就相当于求一个最大生成树的权值最小的那条边，不过用生成树的方法不是超时就RE,只好用最短路中的一个算法解决了，我们看，既然dijkstra是求单源的最短路，d[i]存储的是从起始点到i点的最短路，那么我们就用它来存起始点到i点权值最小的那个；这样答案不就出来了；想想看，这道题逻辑性很强，求能承载的最大重量实际上是求整个联通图中权值最小的，就像短板原理--能装多少水取决于最短的那块木板；
Kruskal生成树：RE，数组开大了又会超时；
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1000+10;
struct node
{
int u,v,w;
}a[N];
int n,m,f[N];
int find(int x)
{
return f[x]==-1?x:f[x]=find(f[x]);
}
int cmp(node a,node b)
{
return a.w>b.w;
}
int ks(int n,int m)
{
for(int i=0;i<m;i++)
scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].w);
sort(a,a+m,cmp);
int minn=10000000;
memset(f,-1,sizeof(f));
for(int i=0;i<m;i++)
{
int x=find(a[i].u);
int y=find(a[i].v);
if(x!=y)
{
minn=min(minn,a[i].w);
f[x]=y;
}
if(find(1)==find(n))
break;
}
return minn;
}
int main()
{
int t;
scanf("%d",&t);
int t1=t;
while(t--)
{
memset(a,0,sizeof(a));
scanf("%d%d",&n,&m);
int x=ks(n,m);
printf("Scenario #%d:\n%d\n",t1-t,x);
}
return 0;
}
Dijkstra最短路变形：AC
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1005;
int w[N][N],v[N],d[N],n,m,t,t1;
void dijkstra()
{
int a,b,c,i,j;
memset(w,0,sizeof(w));
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
w[a][b]=w[b][a]=c;
}
memset(v,0,sizeof(v));
for(i=1; i<=n; i++)
d[i]=w[1][i];
for(i=1;i<=n;i++)
{
int x,m=-1;
for(j=1;j<=n;j++)
if(!v[j]&&d[j]>m)
m=d[x=j];
v[x]=1;
for(j=1;j<=n;j++)
if(!v[j] && d[j]<min(d[x],w[x][j]))
d[j]=min(d[x],w[x][j]);
}
printf("Scenario #%d:\n",t1-t);
printf("%d\n\n",d[n]);
}
int main()
{
scanf("%d",&t);
t1=t;
while(t--)
{
scanf("%d%d",&n,&m);
dijkstra();
}
return 0;
}