POJ 1797 Heavy Transportation（Dijkstra运用）

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

题意：这个题目呢，跟上一篇的题意有点相反，这一次是求在街道1到达街道n的路径所能承受的最大重量。也就是求能从1到达n的路径上的最小承重的最大值。
思路：Dijkstra运用，我们知道dijkstra是每一次将离源点最近的那一一个点进行松弛，而我们现在要求最小承重的最大值，那我们就应该将离源点承重最大的那个点进行松弛。

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 int n, m, dis[1010], mp[1010][1010], vis[1010];
 7 void Dijkstra()
 8 {
 9     for (int i = 1; i <= n; i++) {
10         vis[i] = 0; dis[i] = mp[1][i];//初始化为1到i的最大承重
11     }
12     for (int i = 1; i <= n; i++) {
13         int cnt = 0, k;
14         for (int j = 1; j <= n; j++) {
15             if (!vis[j] && dis[j] > cnt) {
16                 cnt = dis[j];
17                 k = j;
18             }
19         }
20         vis[k] = 1;
21         for (int j = 1; j <= n; j++) {
22             if (!vis[j] && dis[j] < min(dis[k], mp[k][j]))
23                 dis[j] = min(dis[k], mp[k][j]);
24         }
25     }
26 }
27 int main()
28 {
29     ios::sync_with_stdio(false);
30     int T;
31     cin >> T;
32     for(int t=1;t<=T;t++){
33         cin >> n >> m;
34         memset(mp, 0, sizeof(mp));
35         for (int a, b, c, i = 0; i < m; i++) {
36             cin >> a >> b >> c;
37             mp[a][b] = mp[b][a] = c;
38         }
39         Dijkstra();
40         cout << "Scenario #" << t << ":" << endl;
41         cout << dis[n] << endl << endl;
42     }
43     return 0;
44 }