POJ - 1797 Heavy Transportation

Posted 2020-10-04 euzmin

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

这题题意理解很重要，大致意思是给出每条路的载重，求每条路连通且总载重最大时的最小边的值。
（在这里吐槽下网上说的最小值的最大值，看了半天都不理解，我还专门去查了下博弈论和唯物辩证法，都没有关于最小值的最大值的解释，希望以后能懂）
这题我用的是kruskal求的最大生成树（其实就是把从小到大排序变成从大到小排序了），这题还可以用spfa和dijkstra，这里就不尝试了。

好了不废话了，附代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <queue>
 5 using namespace std;
 6 const int inf = 0x3f3f3f3f;
 7 const int M = 1005;
 8 struct nod{
 9     int u,v,cost;
10 }eg[M*M/2];
11 int V,E;
12 bool cmp(const nod&a,const nod&b){
13     return a.cost>b.cost;
14 }
15 int rk[M],pre[M];
16 void init(int V){
17     for(int i=0;i<=V;i++){
18         rk[i]=0;
19         pre[i]=i;
20     }
21 }
22 int find(int x){
23     if(pre[x]==x) {
24         return x;
25     } else{
26         return pre[x]=find(pre[x]);
27     }
28 }
29 void mix(int x,int y){
30     x=find(x);
31     y=find(y);
32     if(x==y) return ;
33     if(rk[x]<y){
34         pre[x]=y;
35     } else{
36         pre[y]=x;
37         if(rk[x]==rk[y]){
38             rk[x]++;
39         }
40     }
41 }
42 int kruskal(){
43     sort(eg,eg+E,cmp);
44   //  printf("%d\n",E);
45     int minn=inf;
46     for(int i=0;i<E;i++){
47         nod e=eg[i];
48         if(find(e.u)!=find(e.v)){
49             mix(e.u,e.v);
50             if(find(1)==find(V)){
51                 minn=e.cost;
52                 break;
53             }
54         }
55     }
56     return minn;
57 }
58 int main(){
59     int t;
60     int cas=0;
61     scanf("%d",&t);
62     while(t--){
63         scanf("%d %d",&V,&E);
64         init(V);
65        // int u,v,cost;
66         for(int i=0;i<E;i++){
67             scanf("%d %d %d",&eg[i].u,&eg[i].v,&eg[i].cost);
68         }
69         printf("Scenario #%d:\n%d\n\n",++cas,kruskal());
70 
71     }
72     return 0;
73 }

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