POJ1797-Heavy Transportation（Dijkstra 变式& 最大生成树）
Heavy Transportation
Time Limit: 3000MS Memory Limit: 30000K
Total Submissions: 32337 Accepted: 8567
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
分析
两种方法，一种是Dijkstra变式，一种是最大生成树，在最大生成树路径中找权值最小的边。
Case 1:
Dijkstra 变式(这题与Frog 不同，Frog 是求通路中 希望每条边尽可能小，这题是希望通路中每条边尽可能大)
#includescanf("%d%d%d",&t1,&t2,&t3);
g[t1][t2]=g[t2][t1]=t3;
}
init(n);
dijkstra(n);
printf("Scenario #%d:\n",cas);
printf("%d\n\n",dist[n]);
}
}
Case 2:
最大生成树 & kruskal , 但是当起点与终点 在一个集合时要终止，否则会受到后续边的影响。
#include1;i++){
if(find(edge[i].st) != find(edge[i].ed)){
Union(edge[i].st,edge[i].ed);
rst--;
if(find(1) == find(n) )
return edge[i].cost;
}
}
return -1;
}
int main(){
// freopen("in.txt","r",stdin);
int T;
scanf("%d",&T);
for(int cas=1;cas <=T;cas ++){
int n,m;
scanf("%d%d",&n,&m);
int t1,t2,t3;
for(int i=0;iscanf("%d%d%d",&t1,&t2,&t3);
edge[i].st=t1;
edge[i].ed=t2;
edge[i].cost =t3;
}
init(n);
printf("Scenario #%d:\n",cas);
printf("%d\n\n",kruskal(n,m) );
}
}